If a resistor has a resistance of 10 Ω and the potential difference across it is 20 V, what is the current flowing through it?

Correct Option: C. Solution: Using Ohm's Law, V = IR, we find I = V/R = 20 V / 10 Ω = 2 A.

In the circuit described, if the equivalent resistance $R_{eq}$ is $(5/6) \, \Omega$ and the emf is 10 V, what is the total current flowing in the network?

Correct Option: C. Solution: Using Ohm's law, $I = V/R_{eq} = 10 \, \text{V} / (5/6) \, \Omega = 12 \, \text{A}$.

If a circuit consists of a 10 V battery and a resistor of 5 Ω, what is the current flowing through the circuit?

Correct Option: B. Solution: Using Ohm's law, $V = IR$, the current $I = \frac{V}{R} = \frac{10}{5} = 2$ A.

Current Electricity

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CBSE Learning Objectives – Key Concepts & Skills You Must Know

Learning Objectives

Understand and apply Kirchhoff's Junction Rule in circuit analysis.
Understand and apply Kirchhoff's Loop Rule in circuit analysis.
Analyze the Wheatstone bridge and determine unknown resistances.
Define and calculate electric current, voltage, resistance, and resistivity.
Explain the limitations of Ohm's Law and identify materials that do not obey it.
Calculate current density and drift velocity in conductive materials.
Describe the relationship between electric field, drift velocity, and current density.
Identify the factors affecting resistance in conductors and their implications in circuit design.

CBSE Revision Notes & Quick Summary for Last-Minute Study

Chapter Notes on Current Electricity

1. Kirchhoff's Rules

Junction Rule: At any junction of circuit elements, the sum of currents entering the junction must equal the sum of currents leaving it.
Loop Rule: The algebraic sum of changes in potential around any closed loop must be zero.

2. Wheatstone Bridge

Arrangement of four resistances: R₁, R₂, R₃, R₄.
Null-point condition: $R_1 / R_2 = R_3 / R_4$

3. Physical Quantities and Units

Physical Quantity	Symbol	Dimensions	Unit	Remark
Electric current	I	[A]	A	SI base unit
Charge	Q,q	[T A]	C
Voltage, Electric potential difference	V	[ML²T³A⁻¹]	V	Work/charge
Electromotive force	ε	[M L² T⁻³ A⁻¹]	V	Work/charge
Resistance	R	[ML²T⁻³A⁻²]	C	R = V/I
Resistivity	p	[ML³T⁻³A⁻²]	Ω m	R = pl/A
Electrical conductivity	σ	[M⁻¹ L⁻³ T³ A²]	S	σ = 1/p
Electric field	E	[M LT ³ A⁻¹]	V m⁻¹	Electric force charge
Drift speed	v_d	[LT⁻¹]	m s⁻¹	v_a = et
Relaxation time	τ	[T]	S
Current density	j	[L⁻² A]	A m⁻²	current/area
Mobility	µ		m²V⁻¹s⁻¹	v_a/E

4. Current and Charge Flow

Current is a scalar quantity, represented by an arrow.
The current I through an area of cross-section is given by the scalar product: $I = j S$

5. Ohm's Law and Limitations

Ohm's Law: $V = IR$ is not universally applicable.
Limitations include:
- V is not proportional to I.
- Relation between V and I depends on the sign of V.
- Non-unique relation between V and I.

6. Important Concepts

Drift Velocity: The average velocity of charge carriers in a conductor due to an electric field.
Resistivity: Depends on material and temperature.
Current Density: Amount of charge flowing per second per unit area.

7. Kirchhoff's Junction Rule

Based on conservation of charge; outgoing currents equal incoming currents at a junction.

8. Example Circuit Analysis

Example circuit with resistors and voltage sources illustrating Kirchhoff's rules and current flow.

9. Exercises

Problems related to current, resistance, and circuit analysis.

CBSE Exam Tips, Important Questions & Common Mistakes to Avoid

Common Mistakes and Exam Tips

Common Pitfalls

Misunderstanding Ohm's Law: Students often think that the equation V = IR is a fundamental law applicable to all devices. However, this equation defines resistance and may not hold true for non-ohmic devices like diodes.
Ignoring Kirchhoff's Rules: Failing to apply Kirchhoff's junction rule correctly can lead to incorrect current calculations at circuit junctions. Remember that the sum of currents entering a junction must equal the sum of currents leaving it.
Confusing Current and Charge: Current is a scalar quantity, and students may mistakenly treat it as a vector, leading to errors in calculations involving multiple currents.

Tips for Success

Understand the Conditions for Ohm's Law: Recognize that Ohm's law applies only when the resistivity of the material does not depend on the electric field's magnitude and direction.
Practice Circuit Analysis: Regularly practice applying Kirchhoff's rules to various circuit configurations to build confidence and accuracy in solving circuit problems.
Visualize Current Flow: Draw diagrams of circuits and label all currents and voltages clearly. This will help in understanding the relationships between different components in the circuit.
Review Characteristics of Materials: Familiarize yourself with the properties of conductors, semiconductors, and insulators, as their behavior under electric fields can differ significantly.

CBSE Quiz & Practice Test – MCQs, True/False Questions with Solutions

Multiple Choice Questions

4.25 Ω

0.0425 Ω

0.000425 Ω

0.425 Ω

Correct Answer: B

Solution:

The resistance

R

is given by

R = \frac{\rho l}{A} = \frac{1.7 \times 10^{-8} \times 5}{0.02} = 0.0425 \, \Omega

4.8 V

8 V

9.6 V

10 V

Correct Answer: C

Solution:

The equivalent emf for cells in parallel is given by

\varepsilon_{eq} = \frac{\varepsilon_1 r_2 + \varepsilon_2 r_1}{r_1 + r_2}

. Substituting the given values,

\varepsilon_{eq} = \frac{12 \times 3 + 6 \times 2}{2 + 3} = 9.6 \text{ V}

1 A

2 A

0.5 A

1.5 A

Correct Answer: C

Solution:

Applying Kirchhoff's loop rule, the total voltage around the loop must be zero. Let the current through the resistor connected to the battery be I. The voltage across the other two resistors will be 2I each. Thus, 6 V - 2I - 2I = 0, solving gives I = 0.5 A.

Zero

Equal to the current in one of the resistors

Double the current in one of the resistors

Half the current in one of the resistors

Correct Answer: A

Solution:

In a balanced Wheatstone bridge, the potential difference across the galvanometer is zero, so no current flows through it.

7.5 V

6.67 V

5 V

8 V

Correct Answer: B

Solution:

The equivalent emf

\varepsilon_{\text{eq}}

for cells in parallel is given by

\varepsilon_{\text{eq}} = \frac{\varepsilon_1 r_2 + \varepsilon_2 r_1}{r_1 + r_2} = \frac{10 \times 2 + 5 \times 1}{1 + 2} = \frac{20 + 5}{3} = 6.67\, \text{V}

R = \rho \frac{l}{A}

R = \frac{A}{\rho l}

R = \frac{\rho A}{l}

R = \frac{\rho l}{A}

Correct Answer: D

Solution:

The resistance

R

of a conductor is given by

R = \rho \frac{l}{A}

, where

\rho

is the resistivity.

R = \rho \frac{l}{A}

R = \rho \frac{A}{l}

R = \frac{l}{\rho A}

R = \frac{A}{\rho l}

Correct Answer: A

Solution:

The resistance

R

of a conductor is given by

R = \rho \frac{l}{A}

, where

\rho

is the resistivity,

l

is the length, and

A

is the cross-sectional area.

25 Ω

100 Ω

200 Ω

50 Ω

Correct Answer: B

Solution:

For a balanced Wheatstone bridge, the ratio R1/R2 = R3/R4. Substituting the given values, 100/200 = 50/R4. Solving for R4 gives R4 = 100 Ω.

1 A

2 A

3 A

4 A

Correct Answer: A

Solution:

The total resistance in the circuit is the sum of the resistances:

R_{total} = 5 \ \Omega + 10 \ \Omega + 15 \ \Omega = 30 \ \Omega

. Using Ohm's Law,

I = \frac{V}{R} = \frac{30 \ \text{V}}{30 \ \Omega} = 1 \ \text{A}

It remains the same.

It doubles.

It halves.

It quadruples.

Correct Answer: B

Solution:

The current density

j

is given by

j = nevd

, where

n

is the number density,

e

is the charge of an electron, and

vd

is the drift velocity. If

vd

is doubled, then

j

also doubles.

1.6 A/m²

16 A/m²

160 A/m²

0.16 A/m²

Correct Answer: C

Solution:

The current density

j

is calculated as

j = (5 \times 10^{28} \, m^{-3})(1.6 \times 10^{-19} \, C)(2 \times 10^{-4} \, m/s) = 160 \, A/m^2

0.3 Ω

7.5 Ω

3 Ω

15 Ω

Correct Answer: C

Solution:

The resistance of a conductor is given by R = ρL/A, where ρ is the resistivity, L is the length, and A is the cross-sectional area. Substituting the given values, R = (1.5 Ω·m)(10 m) / (2 m²) = 7.5 Ω / 2 = 3 Ω.

It doubles

It halves

It remains the same

It quadruples

Correct Answer: A

Solution:

The current density

j

is directly proportional to the number density

n

of charge carriers. If

n

is doubled,

j

also doubles.

0.5 A

1 A

1.5 A

2 A

Correct Answer: B

Solution:

The current through the resistor directly connected to the voltage source can be calculated using Ohm's law:

I = \frac{V}{R} = \frac{15 \text{ V}}{10 \Omega} = 1 \text{ A}

R_{AB} \cdot R_{CD} = R_{BC} \cdot R_{DA}

R_{AB} \cdot R_{BC} = R_{CD} \cdot R_{DA}

R_{AB} \cdot R_{DA} = R_{BC} \cdot R_{CD}

R_{AB} + R_{CD} = R_{BC} + R_{DA}

Correct Answer: A

Solution:

For a Wheatstone bridge to be balanced, the product of the resistances in one pair of opposite branches must equal the product of the resistances in the other pair:

R_{AB} \cdot R_{CD} = R_{BC} \cdot R_{DA}

\frac{R_1}{R_2} = \frac{R_3}{R_4}

\frac{R_1}{R_3} = \frac{R_2}{R_4}

\frac{R_2}{R_1} = \frac{R_4}{R_3}

\frac{R_1}{R_4} = \frac{R_2}{R_3}

Correct Answer: C

Solution:

For a Wheatstone bridge to be balanced, the ratio

\frac{R_2}{R_1} = \frac{R_4}{R_3}

must hold true.

2 A

3 A

5 A

8 A

Correct Answer: A

Solution:

According to Kirchhoff's junction rule, the sum of currents entering a junction must equal the sum of currents leaving the junction. Therefore, the current leaving node C must be 5 A (entering) - 3 A (leaving B) = 2 A.

2.56 x 10⁶ A/m²

2.56 x 10⁵ A/m²

2.56 x 10³ A/m²

2.56 x 10⁴ A/m²

Correct Answer: A

Solution:

The current density j is given by j = nqv_d, where n is the electron density, q is the charge of an electron, and v_d is the drift velocity. Substituting the given values, j = (8 x 10²⁸ electrons/m³)(1.6 x 10⁻¹⁹ C)(2 x 10⁻⁴ m/s) = 2.56 x 10⁶ A/m².

10 V

5 V

0 V

20 V

Correct Answer: A

Solution:

The potential difference across a resistor directly connected to a 10 V source is 10 V.

Resistor

Diode

Capacitor

Inductor

Correct Answer: B

Solution:

A diode is a non-linear device where the current-voltage relationship is not proportional, especially when the diode is forward or reverse biased. This behavior is typical in semiconductor devices.

1 Ω

2 Ω

\frac{5}{6} , \Omega

\frac{3}{2} , \Omega

Correct Answer: C

Solution:

The network is symmetric, and using Kirchhoff's rules, the equivalent resistance can be calculated as

R_{eq} = \frac{5}{6} \, \Omega

0.01 Ω

0.1 Ω

1 Ω

10 Ω

Correct Answer: A

Solution:

The resistance

R

of a conductor is given by

R = \frac{\rho l}{A}

, where

\rho

is the resistivity,

l

is the length, and

A

is the cross-sectional area. Substituting the given values,

R = \frac{1 \times 10^{-8} \ \Omega \cdot m \times 1 \ m}{1 \times 10^{-4} \ m^2} = 0.01 \ \Omega

2 A

0.5 A

5 A

10 A

Correct Answer: A

Solution:

Using Ohm's law,

V = IR

, where

V = 10 \, \text{V}

and

R = 5 \, \Omega

. Solving for

I

gives

I = V/R = 10/5 = 2 \, \text{A}

I

2I

4I

5I

Correct Answer: D

Solution:

The total current is the sum of all currents flowing out of the voltage source, which is

I + 3I + I/2 = 5I

j = I / (πR²)

j = nqv_d

j = I / (2πR)

j = nqv_d / A

Correct Answer: B

Solution:

The current density j is defined as the current per unit area. For a cylindrical conductor, j = nqv_d, where n is the number density of charge carriers, q is the charge of an electron, and v_d is the drift velocity.

10 V

5 V

2 V

1 V

Correct Answer: A

Solution:

Using Ohm's Law, V = IR, the voltage across the resistor is 2 A * 5 Ω = 10 V.

5 V

10 V

20 V

30 V

Correct Answer: C

Solution:

Using Ohm's law, V = IR, where I = 2 A and R = 10 Ω. Therefore, V = 2 A * 10 Ω = 20 V.

It is halved.

It remains the same.

It is doubled.

It is quadrupled.

Correct Answer: C

Solution:

The current density

j

is directly proportional to the number density

n

. Therefore, if the number density is doubled, the current density will also be doubled.

The temperature coefficient is negative.

The temperature coefficient is positive.

The temperature coefficient is zero.

The temperature coefficient is undefined.

Correct Answer: B

Solution:

The temperature coefficient of resistivity is defined as the fractional increase in resistivity per unit increase in temperature. If resistivity increases with temperature, the coefficient is positive.

The current is directly proportional to voltage.

The current is inversely proportional to voltage.

The current does not change with voltage.

The current depends on the direction of the applied voltage.

Correct Answer: D

Solution:

Diodes exhibit a non-linear I-V characteristic where the current depends on the direction of the applied voltage, allowing current to flow primarily in one direction.

0.5 A

1 A

2 A

4 A

Correct Answer: C

Solution:

Using Ohm's Law, V = IR, we find I = V/R = 20 V / 10 Ω = 2 A.

1 Ω

5/6 Ω

1.5 Ω

2 Ω

Correct Answer: B

Solution:

Using symmetry and Kirchhoff's rules, the equivalent resistance of the cube network is calculated as 5/6 Ω.

2 A

3 A

5 A

8 A

Correct Answer: A

Solution:

According to Kirchhoff's junction rule, the sum of currents entering a junction equals the sum of currents leaving. Therefore, if 5 A enters and 3 A leaves through one branch, the other branch must carry 5 A - 3 A = 2 A.

To measure the total voltage across the circuit.

To measure the current through the entire circuit.

To detect the presence of current between nodes B and D.

To provide additional resistance to the circuit.

Correct Answer: C

Solution:

The galvanometer is used to detect the presence of current between nodes B and D, indicating whether the bridge is balanced.

6.67 Ω

15 Ω

30 Ω

5 Ω

Correct Answer: A

Solution:

The equivalent resistance for resistors in parallel is given by

\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{10} + \frac{1}{20} = \frac{3}{20}

. Thus,

R_{eq} = \frac{20}{3} = 6.67 \, \Omega

I/2

3I/2

I/4

Correct Answer: A

Solution:

By the junction rule, the current entering a junction equals the current leaving. Therefore, the current leaving node C is I.

1.2 A

1.0 A

2.0 A

0.8 A

Correct Answer: A

Solution:

The total resistance in the circuit is the sum of the two resistors: 4 Ω + 6 Ω = 10 Ω. Using Ohm's Law, V = IR, the current I is given by I = V/R = 12 V / 10 Ω = 1.2 A.

6.25 \times 10^{-4} \, \text{m/s}

1.6 \times 10^{-3} \, \text{m/s}

10^{-3} \, \text{m/s}

10^{-5} \, \text{m/s}

Correct Answer: A

Solution:

The drift velocity

v_d

is given by

j = ne v_d

. Solving for

v_d

, we get

v_d = \frac{j}{ne} = \frac{10^6}{10^{28} \times 1.6 \times 10^{-19}} = 6.25 \times 10^{-4} \, \text{m/s}

6 A

10 A

12 A

15 A

Correct Answer: C

Solution:

Using Ohm's law,

I = V/R_{eq} = 10 \, \text{V} / (5/6) \, \Omega = 12 \, \text{A}

The current remains the same

The current doubles

The current reverses direction

The current becomes zero

Correct Answer: C

Solution:

If the electric field is reversed, the direction of the force on the charge carriers is reversed, causing the current to reverse direction.

10 W

20 W

5 W

2 W

Correct Answer: B

Solution:

The power dissipated in the resistor is given by

P = \frac{V^2}{R} = \frac{10^2}{5} = 20 \, \text{W}

V - I_1 R_1 - I_2 R_2 - I_3 R_3 = 0

V + I_1 R_1 + I_2 R_2 + I_3 R_3 = 0

V = I_1 R_1 + I_2 R_2 + I_3 R_3

V = I_1 R_1 - I_2 R_2 + I_3 R_3

Correct Answer: A

Solution:

According to Kirchhoff's loop rule, the algebraic sum of potential differences in any closed loop is zero. Thus,

V - I_1 R_1 - I_2 R_2 - I_3 R_3 = 0

100/10 = 5/60

100/60 = 10/5

100/5 = 60/10

100/10 = 60/5

Correct Answer: D

Solution:

For a Wheatstone bridge to be balanced, the ratio of resistances in one branch must equal the ratio in the other branch:

\frac{100}{10} = \frac{60}{5}

R_{eq} = R_1 + R_2

R_{eq} = \frac{R_1 R_2}{R_1 + R_2}

R_{eq} = \frac{R_1 + R_2}{R_1 R_2}

R_{eq} = \frac{1}{R_1} + \frac{1}{R_2}

Correct Answer: B

Solution:

For resistors in parallel, the equivalent resistance is given by

R_{eq} = \frac{R_1 R_2}{R_1 + R_2}

0.5 A

1 A

2 A

5 A

Correct Answer: B

Solution:

Using Ohm's law,

V = IR

, the current

I = \frac{V}{R} = \frac{10}{5} = 2

15 V

12 V

9 V

3 V

Correct Answer: A

Solution:

When cells are connected in series, the total emf is the sum of the individual emfs:

\text{Total emf} = 9 \text{ V} + 6 \text{ V} = 15 \text{ V}

Correct Answer: C

Solution:

For the Wheatstone bridge to be balanced, the ratio R1/R2 must equal R3/R4. Here, R1/R2 = 50/100 = 0.5, and R3/R4 = 25/50 = 0.5. The bridge is already balanced, but if any resistor must be adjusted, it would typically be R3 to maintain the balance if other resistors change.

Glass

Copper

Silicon

Rubber

Correct Answer: B

Solution:

Metals like copper have low resistivity, typically in the range of

10^{-8} \, \Omega \cdot m

10^{-6} \, \Omega \cdot m

43 mA

45 mA

47 mA

50 mA

Correct Answer: C

Solution:

The voltage across the resistor is

5 - 0.7 = 4.3\, \text{V}

. Using Ohm's Law, the current

I

I = \frac{V}{R} = \frac{4.3}{100} = 0.043\, \text{A} = 43\, \text{mA}

The current remains the same.

The current decreases.

The current increases.

The current becomes zero.

Correct Answer: D

Solution:

Diodes allow current to flow easily in one direction (forward bias) but block current in the reverse direction (reverse bias), hence the current becomes zero when the voltage is reversed.

I

2I

I/2

3I

Correct Answer: A

Solution:

The total current entering the segment is

I

, which then splits equally into two branches carrying

I/2

each.

2I/3

I/2

I/4

I

Correct Answer: A

Solution:

According to Kirchhoff's junction rule, the total current entering a junction equals the total current leaving. Therefore, the current in the other branch is

I - I/3 = 2I/3

The current doubles.

The current halves.

The current remains the same.

The current becomes zero.

Correct Answer: A

Solution:

Ohm's law states that

V = IR

. If

V

is doubled and

R

remains constant, then

I

must also double.

The current remains

I

in each outgoing branch.

The current splits equally into two outgoing branches.

The current doubles in each outgoing branch.

The current is halved in each outgoing branch.

Correct Answer: B

Solution:

At corners A', B, and D, the incoming current

I

splits equally into the two outgoing branches.

j = nqv_d

j = qnv_d

j = v_d/nq

j = qv_d/n

Correct Answer: A

Solution:

The current density

j

is given by

j = nqv_d

, where

n

is the number density of charge carriers,

q

is the charge of each carrier, and

v_d

is the drift velocity.

I = V/R

I = VR

I = R/V

I = V^2/R

Correct Answer: A

Solution:

Ohm's Law states that

V = IR

, which can be rearranged to find the current:

I = V/R

The junction rule states that the sum of the currents entering a junction is equal to the sum of the currents leaving the junction.

The loop rule states that the sum of the resistances in a closed loop is zero.

Kirchhoff's rules are only applicable to series circuits.

Kirchhoff's rules can be used to determine the power dissipated in a circuit.

Correct Answer: A

Solution:

Kirchhoff's junction rule states that the sum of the currents entering a junction is equal to the sum of the currents leaving the junction. This is a consequence of the conservation of charge.

2 A

3 A

5 A

8 A

Correct Answer: A

Solution:

According to Kirchhoff's junction rule, the sum of currents entering a junction equals the sum of currents leaving. Therefore, the remaining current must be 2 A.

Ohm's Law

Kirchhoff's Loop Rule

Kirchhoff's Junction Rule

Faraday's Law

Correct Answer: B

Solution:

Kirchhoff's Loop Rule states that the sum of the potential differences (voltage) around any closed loop in a circuit must be zero.

1 Ω

2 Ω

5/6 Ω

3 Ω

Correct Answer: C

Solution:

Using symmetry and Kirchhoff's rules, the equivalent resistance is calculated as

R_{eq} = \frac{5}{6} \Omega

8 V

9 V

10 V

11 V

Correct Answer: A

Solution:

Using the formula V_ext = ε - Ir, where ε is the emf, I is the current, and r is the internal resistance. The total resistance is 1 Ω + 4 Ω = 5 Ω, so I = 10 V / 5 Ω = 2 A. Therefore, V_ext = 10 V - (2 A * 1 Ω) = 8 V.

1 Ω

2 Ω

3 Ω

0.5 Ω

Correct Answer: C

Solution:

The equivalent resistance of resistors in series is the sum of their resistances: 1 Ω + 1 Ω + 1 Ω = 3 Ω.

R_1/R_2 = R_3/R_4

R_1/R_3 = R_2/R_4

R_1/R_4 = R_2/R_3

R_1/R_2 = R_4/R_3

Correct Answer: B

Solution:

For a Wheatstone bridge to be balanced, the ratio of the resistances in one branch must equal the ratio in the other branch:

R_1/R_3 = R_2/R_4

0.5 A

1 A

1.5 A

2 A

Correct Answer: A

Solution:

The total resistance in the series circuit is

R_{total} = R_1 + R_2 + R_3 = 4 + 6 + 8 = 18 \, \Omega

. Using Ohm's Law,

I = \frac{V}{R_{total}} = \frac{12}{18} = 0.5 \, \text{A}

I/2

I

2I

0

Correct Answer: A

Solution:

According to Kirchhoff's junction rule, the current entering a junction equals the total current leaving. If

I

splits equally, each branch carries

I/2

1.5 A

2 A

3 A

4 A

Correct Answer: A

Solution:

The total resistance in the circuit is the sum of the resistances:

R_{\text{total}} = 2 + 3 + 5 = 10\, \Omega

. Using Ohm's Law, the current

I

is given by

I = \frac{V}{R} = \frac{15}{10} = 1.5\, \text{A}

Metals have high resistivity.

Insulators have low resistivity.

Semiconductors have resistivity between metals and insulators.

All materials have the same resistivity.

Correct Answer: C

Solution:

Semiconductors like Si and Ge have resistivity values that are between those of metals and insulators.

1:2

2:3

3:4

1:1

Correct Answer: A

Solution:

For a balanced Wheatstone bridge, the ratio of resistances in one arm is equal to the ratio in the other arm:

\frac{R_1}{R_2} = \frac{R_3}{R_4}

. Given

R_1 = 10 \ \Omega, R_2 = 20 \ \Omega, R_3 = 30 \ \Omega, R_4 = 40 \ \Omega

, the ratio is

\frac{10}{20} = \frac{30}{40} = \frac{1}{2}

0.03 Ω

0.05 Ω

0.02 Ω

0.01 Ω

Correct Answer: A

Solution:

Using the formula R = \rho \frac{l}{A}, where \rho = 1.5 x 10^-8 Ωm, l = 2 m, and A = 1 x 10^-6 m², the resistance R = 1.5 x 10^-8 Ωm * \frac{2 m}{1 x 10^-6 m²} = 0.03 Ω.

2 A

3 A

5 A

8 A

Correct Answer: A

Solution:

According to the junction rule, the sum of currents entering a junction equals the sum of currents leaving. Therefore, if 5 A enters at A and 3 A leaves at B, 2 A must leave at C.

1 A

5 A

2 A

3 A

Correct Answer: B

Solution:

According to Kirchhoff's junction rule, the sum of currents entering a junction equals the sum of currents leaving. Therefore,

I_3 = I_1 + I_2 = 3 \, \text{A} + 2 \, \text{A} = 5 \, \text{A}

1.2 A

2.4 A

0.8 A

1.0 A

Correct Answer: A

Solution:

The total resistance in the series circuit is R_total = R1 + R2 + R3 = 2 Ω + 3 Ω + 5 Ω = 10 Ω. Using Ohm's Law, V = IR, the current I = V/R_total = 12 V / 10 Ω = 1.2 A.

15 V

10 V

5 V

20 V

Correct Answer: A

Solution:

Using Ohm's Law,

V = IR = 3 \, A \times 5 \, \Omega = 15 \, V

15 Ω

20 Ω

25 Ω

30 Ω

Correct Answer: A

Solution:

In a balanced Wheatstone bridge, the ratio of resistances in one branch is equal to the ratio in the other branch. Therefore, R1/R2 = R3/R4. Solving gives the resistance across the bridge as 15 Ω.

Copper

Silicon

Glass

Rubber

Correct Answer: B

Solution:

Semiconductors like silicon have resistivities in the range of

10^{-4} \, \Omega \cdot m

, which is much higher than metals but lower than insulators.

I_1 + I_2 = I_3

I_1 = I_2 + I_3

I_1 + I_3 = I_2

I_1 = I_2 - I_3

Correct Answer: B

Solution:

Kirchhoff's junction rule states that the sum of currents entering a junction equals the sum of currents leaving. Thus,

I_1 = I_2 + I_3

True or False

Correct Answer: True

Solution:

Current density

j

is given by

j = nq v_d

, where

n

is the number density of charge carriers,

q

is the charge of the carriers, and

v_d

is the drift velocity.

Correct Answer: True

Solution:

This is known as Kirchhoff's junction rule, which states that the total current entering a junction equals the total current leaving the junction.

Correct Answer: False

Solution:

The resistance of a conductor depends on both its material (resistivity) and its dimensions (length and cross-sectional area).

Correct Answer: False

Solution:

The resistance of a conductor depends on both the material and its dimensions, specifically its length and cross-sectional area.

Correct Answer: False

Solution:

The resistivity of a conductor is a material property and does not depend on its length or cross-sectional area. However, the resistance of a conductor depends on these dimensions.

Correct Answer: True

Solution:

The excerpt mentions labeled nodes in a circuit diagram, including points A, A', B, B', C, C', D, D', and E.

Correct Answer: True

Solution:

The circuit described has symmetrical paths, which implies that the current splits equally at each junction, as indicated by the symmetry in the problem.

Correct Answer: True

Solution:

Kirchhoff's junction rule is a fundamental principle in circuit analysis, ensuring the conservation of charge at a junction.

Correct Answer: True

Solution:

In a balanced Wheatstone bridge, the potential difference across the galvanometer is zero, resulting in no current flow through it.

Correct Answer: False

Solution:

Not all networks can be reduced to simple series and parallel combinations. Some require the application of Kirchhoff's rules to solve.

Correct Answer: True

Solution:

Electrons, being negatively charged, move opposite to the direction of the electric field.

Correct Answer: True

Solution:

The symmetry in a resistor network can allow for simplifications in calculating equivalent resistance, as shown in the example where paths AA', AD, and AB are symmetrically placed.

Correct Answer: False

Solution:

Ohm's law is not a fundamental law of nature. It does not apply to all materials, especially those where the relationship between voltage and current is non-linear.

Correct Answer: True

Solution:

The diagram indicates the current

I

flowing through the segment, which means the actual current is

I

unless otherwise specified.

Correct Answer: True

Solution:

This statement is true as it follows Kirchhoff's junction rule, which states that at any junction, the sum of the currents entering the junction is equal to the sum of currents leaving the junction.

Correct Answer: False

Solution:

The network is not reducible to a simple series and parallel combinations of resistors. Symmetry and Kirchhoff's rules are used to determine the equivalent resistance.

Correct Answer: True

Solution:

The excerpt describes Kirchhoff's junction rule, which states that the sum of currents entering a junction is equal to the sum of currents leaving the junction.

Correct Answer: False

Solution:

The resistance

R

of a conductor depends on its length

l

and cross-sectional area

A

through the relation

R = \rho \frac{l}{A}

, where

\rho

is the resistivity.

Correct Answer: True

Solution:

The excerpt explains that when

n

cells are connected in parallel, they can be represented by a single cell with an equivalent emf and internal resistance.

Correct Answer: True

Solution:

Kirchhoff's loop rule is based on the principle that electric potential is dependent on location, so the total change in potential around a closed loop must be zero.

Correct Answer: True

Solution:

In a balanced Wheatstone bridge, the potential difference across the galvanometer is zero, resulting in no current flow through it.

Correct Answer: False

Solution:

The voltage source is indeed located at point E and is marked as 10 V.

Correct Answer: True

Solution:

The circuit described shows current flow with segments carrying values like

I/2

I

, and

3I

, which indicates current division at different paths.

Correct Answer: True

Solution:

In electrostatic equilibrium, the electric field inside a conductor is zero because charges redistribute themselves on the surface.

Correct Answer: True

Solution:

This is Kirchhoff's loop rule, which states that the sum of the electromotive forces and potential differences in any closed loop is zero.

Correct Answer: False

Solution:

Kirchhoff's rules are applicable to any circuit, no matter how complex, to analyze currents and potential differences.

Correct Answer: False

Solution:

The excerpt states that the voltage source is located at point E, not D, and is marked as 10 V.

Correct Answer: True

Solution:

According to Kirchhoff's junction rule, the sum of currents entering a junction is equal to the sum of currents leaving the junction.

Correct Answer: True

Solution:

The voltage source marked as 10 V indicates that it provides a potential difference of 10 volts.

Correct Answer: False

Solution:

This statement is false. Some networks, due to their complexity or symmetry, cannot be reduced to simple series and parallel combinations, and require methods like Kirchhoff's rules to determine the equivalent resistance.

Correct Answer: True

Solution:

This statement is true. Metals generally have low resistivity, which falls within the range of

10^{-8}

Ω m to

10^{-6}

Ω m.

Correct Answer: False

Solution:

Metals have low resistivity, whereas insulators have much higher resistivity values.

Correct Answer: False

Solution:

The resistance

R

of a conductor depends on its length

l

and cross-sectional area

A

according to the relation

R = \rho \frac{l}{A}

, where

\rho

is the resistivity.

Correct Answer: True

Solution:

Kirchhoff's rules can be used to determine the equivalent resistance of a network, especially when symmetry is present, simplifying the analysis.

Correct Answer: True

Solution:

The drift velocity is proportional to the electric field, as described by the relation

v_d = \frac{eE\tau}{m}

, where

\tau

is the average collision time.

Correct Answer: False

Solution:

Ohm's law is not a fundamental law of nature and does not hold for all materials, especially those with non-linear V-I characteristics.

Correct Answer: True

Solution:

Ohm's law is expressed as

V = IR

, where

V

is the voltage,

I

is the current, and

R

is the resistance, indicating a direct proportionality between voltage and current.

Correct Answer: False

Solution:

The resistance of a conductor depends on both the material and the dimensions of the conductor.

Correct Answer: True

Solution:

The current through each resistor can vary depending on how they are connected in the circuit, as indicated by the presence of different current values like

I/2

I

, and

3I

in the excerpt.

Correct Answer: True

Solution:

The current density

j

gives the amount of charge flowing per second per unit area normal to the flow, as defined by

j = nq v_d

, where

n

is the number density of charge carriers,

q

is the charge of the carriers, and

v_d

is the drift velocity.

Correct Answer: True

Solution:

In a balanced Wheatstone bridge, the product of the resistances in one diagonal equals the product in the other diagonal, i.e.,

R_1 R_4 = R_2 R_3

. This is a condition for no current through the galvanometer.

Correct Answer: True

Solution:

In a balanced Wheatstone bridge, the ratio of resistances in one branch is equal to the ratio in the other branch, resulting in no current through the galvanometer.

Correct Answer: True

Solution:

The symmetry in the cubical network allows for simplification in calculating the equivalent resistance, and Kirchhoff's rules can be applied to further solve the problem.

Correct Answer: False

Solution:

Ohm's law is not universally valid for all materials. The excerpt mentions that some materials and devices exhibit non-linear relationships between voltage and current.

Correct Answer: True

Solution:

The equations for series and parallel combinations of resistors remain valid regardless of how the terminals of the cells are connected.

Correct Answer: True

Solution:

According to the formula

R = \frac{\rho l}{A}

, where

\rho

is the resistivity,

l

is the length, and

A

is the cross-sectional area.

Correct Answer: True

Solution:

Current density

j

is defined as the amount of charge flowing per second per unit area normal to the flow, given by

j = nq v_d

, where

n

is the number density of charge carriers,

q

is the charge, and

v_d

is the drift velocity.

Correct Answer: True

Solution:

This statement is true. A Wheatstone bridge is balanced when the ratio of resistances in one branch (R1/R2) is equal to the ratio in the other branch (R3/R4).

Correct Answer: True

Solution:

This statement is true and is known as Kirchhoff's junction rule, which states that the sum of currents entering a junction equals the sum of currents leaving the junction.

Correct Answer: True

Solution:

This is known as Kirchhoff's loop rule, which states that the sum of the potential differences (voltage) around any closed loop in a circuit is zero.

Correct Answer: True

Solution:

The diagram features intersecting lines and geometric shapes, indicating possible reflections or refractions, which are concepts related to optics or geometry.

Correct Answer: False

Solution:

Kirchhoff's rules can be applied to any circuit, even if it is not reducible to simple series and parallel combinations. They are particularly useful in complex networks.

Correct Answer: True

Solution:

For a Wheatstone bridge to be balanced, the ratio of resistances in one branch must equal the ratio in the other branch. If this condition is not met, the bridge is unbalanced.

Correct Answer: True

Solution:

According to Kirchhoff's junction rule, the sum of currents entering a junction equals the sum of currents leaving the junction.

Correct Answer: True

Solution:

The balance condition for a Wheatstone bridge is

\frac{R_1}{R_2} = \frac{R_3}{R_4}

, which ensures no current flows through the galvanometer.

Correct Answer: False

Solution:

The resistance of a conductor depends on its length and cross-sectional area, as well as the material's resistivity.

Correct Answer: True

Solution:

In electrostatic equilibrium, the charges within a conductor redistribute themselves such that the electric field inside the conductor is zero.

Correct Answer: False

Solution:

Ohm's law is not a fundamental law of nature and may not hold in certain materials or devices where the relationship between voltage and current is non-linear or depends on the sign of the voltage.

Correct Answer: True

Solution:

Semiconductors like silicon and germanium have resistivity values that are higher than metals but lower than insulators, placing them in an intermediate range.

Correct Answer: True

Solution:

The excerpt states that the resistance of a conductor depends on its length and cross-sectional area according to the relation

R = \rho \frac{l}{A}

, where

\rho

is the resistivity.

Correct Answer: True

Solution:

A Wheatstone bridge is balanced when the ratio of resistances in one branch equals the ratio in the other branch, allowing no current to flow through the galvanometer.

Correct Answer: False

Solution:

Kirchhoff's rules are applicable to complex circuits as well, where series and parallel combinations are not sufficient to determine all currents and potential differences.

Current Electricity

AI Learning Assistant

CBSE Learning Objectives – Key Concepts & Skills You Must Know

Learning Objectives

CBSE Revision Notes & Quick Summary for Last-Minute Study

Chapter Notes on Current Electricity

1. Kirchhoff's Rules

2. Wheatstone Bridge

3. Physical Quantities and Units

4. Current and Charge Flow

5. Ohm's Law and Limitations

6. Important Concepts

7. Kirchhoff's Junction Rule

8. Example Circuit Analysis

9. Exercises

CBSE Exam Tips, Important Questions & Common Mistakes to Avoid

Common Mistakes and Exam Tips

Common Pitfalls

Tips for Success

Multiple Choice Questions

Q1.A cylindrical conductor has a length of 5 m and a cross-sectional area of 0.02 m². If the resistivity of the material is 1.7×10−8 Ω⋅m1.7 \times 10^{-8} \, \Omega\cdot m1.7×10−8Ω⋅m, what is the resistance of the conductor?

Correct Answer: B

Q2.In a circuit, two cells with emfs of 12 V and 6 V are connected in parallel with internal resistances of 2 Ω and 3 Ω respectively. What is the equivalent emf of the combination?

Correct Answer: C

Q3.Consider a circuit with three resistors, each of resistance 2 Ω, connected in a triangle. A battery of 6 V is connected across one of the resistors. Using Kirchhoff's rules, determine the current flowing through the resistor connected to the battery.

Correct Answer: C

Q4.In a Wheatstone bridge circuit, if the resistances are balanced, what is the current through the galvanometer?

Correct Answer: A

Q5.In a hypothetical circuit, two cells with emfs of 10 V and 5 V are connected in parallel with internal resistances of 1 Ω and 2 Ω respectively. What is the equivalent emf of the combination?

Correct Answer: B

Q6.If a conductor has a length lll and cross-sectional area AAA, which of the following correctly represents its resistance RRR?

Correct Answer: D

Q7.If a conductor has a length lll and cross-sectional area AAA, and the resistivity of the material is ρ\rhoρ, what is the resistance RRR of the conductor?

Correct Answer: A

Q8.A Wheatstone bridge is balanced with resistors R1 = 100 Ω, R2 = 200 Ω, and R3 = 50 Ω. What is the value of the fourth resistor R4?

Correct Answer: B

Q9.Consider a circuit with three resistors of resistances 5 Ω, 10 Ω, and 15 Ω connected in series with a battery of emf 30 V and negligible internal resistance. What is the current flowing through the circuit?

Correct Answer: A

Q10.If the drift velocity of electrons in a conductor is doubled, what happens to the current density assuming all other factors remain constant?

Correct Answer: B

Correct Answer: C

Q12.A conductor has a cross-sectional area of 2 m² and a length of 10 m. If the resistivity of the material is 1.5 Ω·m, what is the resistance of the conductor?

Correct Answer: C

Q13.In a circuit, the current density is given by j=ne2τE/mj = ne^2 \tau E/mj=ne2τE/m. If the number density of charge carriers is doubled, what happens to the current density?

Correct Answer: A

Q14.Consider a circuit with three resistors, each of resistance 10 Ω, connected in a triangle. A voltage source of 15 V is connected across one of the resistors. What is the current flowing through the resistor directly connected to the voltage source?

Correct Answer: B

Correct Answer: A

Q16.In a Wheatstone bridge, if the resistors are arranged such that R1=100 ΩR_1 = 100 \, \OmegaR1​=100Ω, R2=10 ΩR_2 = 10 \, \OmegaR2​=10Ω, R3=5 ΩR_3 = 5 \, \OmegaR3​=5Ω, and R4=60 ΩR_4 = 60 \, \OmegaR4​=60Ω, what is the condition for the bridge to be balanced?

Correct Answer: C

Q17.Consider a circuit with three nodes labeled A, B, and C. The current entering node A is 5 A, and the current leaving node B is 3 A. If the circuit is in a steady state, what is the current leaving node C?

Correct Answer: A

Q18.In a cylindrical conductor, the drift velocity of electrons is found to be 2 x 10⁻⁴ m/s. If the electron density is 8 x 10²⁸ electrons/m³ and the charge of an electron is 1.6 x 10⁻¹⁹ C, what is the current density?

Correct Answer: A

Q19.If a voltage source of 10 V is connected at point E in a circuit, what is the potential difference across a resistor connected directly to this source?

Correct Answer: A

Q20.A non-linear device exhibits a current-voltage relationship where the current is not directly proportional to the voltage. Which of the following devices is most likely to exhibit such behavior?

Correct Answer: B

Q21.Consider a circuit with a voltage source of 10 V connected to a network of resistors forming a cube, with each edge having a resistance of 1 Ω. What is the equivalent resistance of the network?

Correct Answer: C

Q22.A cylindrical conductor has a resistivity of 1×10−8 Ω⋅m1 \times 10^{-8} \ \Omega \cdot m1×10−8 Ω⋅m, length 1 m, and cross-sectional area 1×10−4 m21 \times 10^{-4} \ m^21×10−4 m2. What is the resistance of the conductor?

Correct Answer: A

Q23.In a circuit with a voltage source of 10 V and a resistor of 5 Ω, what is the current flowing through the resistor?

Correct Answer: A

Q24.In a circuit with a voltage source of 10 V at point E, if the current flowing through a resistor is III, and the current divides into I/2I/2I/2 and 3I3I3I in different branches, what is the total current flowing out of the voltage source?

Correct Answer: D

Q25.A cylindrical conductor of radius R carries a current I. If the drift velocity of electrons is v_d, what is the expression for the current density j in the conductor?

Correct Answer: B

Q26.In a circuit, if the current through a resistor is 2 A and the resistance is 5 Ω, what is the voltage across the resistor?

Correct Answer: A

Q27.If a conductor has a resistance of 10 Ω and a current of 2 A flows through it, what is the potential difference across the conductor?

Correct Answer: C

Correct Answer: C

Q29.If a material has a resistivity that increases with temperature, which of the following statements is true about its temperature coefficient of resistivity?

Correct Answer: B

Q30.A diode exhibits a non-linear I-V characteristic. Which of the following statements about its behavior is true?

Correct Answer: D

Q31.If a resistor has a resistance of 10 Ω and the potential difference across it is 20 V, what is the current flowing through it?

Correct Answer: C

Q32.What is the equivalent resistance of a network where all resistors are 1 Ω and are symmetrically placed in a cube?