Ohm's Law is given by $ V = IR $. If a conductor has a resistance of 5 Ω and a current of 3 A flows through it, what is the potential difference across the conductor?

Correct Option: A. Solution: Using Ohm's Law, $ V = IR = 3 \, A \times 5 \, \Omega = 15 \, V $.

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.

Current Electricity

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Summary

Chapter Summary: Current Electricity

Kirchhoff's Rules

Junction Rule: The sum of currents entering a junction equals the sum of currents leaving it.
Loop Rule: The algebraic sum of changes in potential around any closed loop must be zero.

Wheatstone Bridge

Arrangement of four resistances: R₁, R₂, R₃, R₄.
Null-point condition: R₁/R₃ = R₂/R₄.

Physical Quantities and Units

Electric Current (I): [A], SI base unit.
Charge (Q, q): [T A], Unit: C.
Voltage (V): [ML²T³A⁻¹], Unit: V (Work/charge).
Electromotive Force (ε): [ML²T⁻³A⁻¹], Unit: V (Work/charge).
Resistance (R): [ML²T⁻³A⁻²], Unit: Ω, R = V/I.
Resistivity (ρ): [ML³T⁻³A⁻²], Unit: Ω m, R = ρl/A.
Electrical Conductivity (σ): [M⁻¹L⁻³T³A²], Unit: S, σ = 1/ρ.
Electric Field (E): [MLT³A⁻¹], Unit: V m⁻¹.
Drift Speed (v_d): [LT⁻¹], Unit: m s⁻¹, v_d = eτ.
Relaxation Time (τ): [T], Unit: S.
Current Density (j): [L⁻²A], Unit: A m⁻², j = I/A.
Mobility (μ): m²V⁻¹s⁻¹, μ = v_a/E.

Important Concepts

Current is a scalar quantity.
Ohm's Law: V = IR, applicable to all conducting devices.
Kirchhoff's rules are based on conservation of charge.

Limitations of Ohm's Law

V may not be proportional to I in certain materials (e.g., diodes).
The relationship between V and I can depend on the direction of V.
Non-unique V for the same I can occur in some materials.

Applications and Examples

Wheatstone Bridge: Used to determine unknown resistances.
Current Density: j = nqv_d, where n is the number density of charge carriers.
Resistivity: Varies with temperature and material type.

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.

Detailed Notes

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.

Exam Tips & Common Mistakes

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.

Practice & Assessment

Multiple Choice Questions

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}

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.

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}

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.

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.

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

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

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.

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}

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

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}

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.

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

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}

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.

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.

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

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.

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.

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.

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.

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}

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

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.

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.

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}

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

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.

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.

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.

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}

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.

1.2 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.

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.

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 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}

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.

\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.

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

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.

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

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}

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.

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

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.

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.

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}

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

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.

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.

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

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.

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.

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.

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}

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.

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.

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.

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.

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.

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.

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}

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.

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}

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}

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.

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

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}

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

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

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.

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.

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

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.

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.

True or False

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:

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:

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: 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:

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: 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.

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:

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:

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:

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

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:

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

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 balance condition for a Wheatstone bridge is

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

l

and cross-sectional area

A

through the relation

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

, where

\rho

is the resistivity.

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:

The diagram indicates the current

I

flowing through the segment, which means the actual current is

I

unless otherwise specified.

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.,

Solution:

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

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:

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

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: True

Solution:

The circuit described shows current flow with segments carrying values like

I/2

I

, and

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:

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

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:

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:

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. It does not apply to all materials, especially those where the relationship between voltage and current is non-linear.

Current Electricity

AI Learning Assistant

Summary

Chapter Summary: Current Electricity

Kirchhoff's Rules

Wheatstone Bridge

Physical Quantities and Units

Important Concepts

Limitations of Ohm's Law

Applications and Examples

Learning Objectives

Learning Objectives

Detailed Notes

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

Exam Tips & Common Mistakes

Common Mistakes and Exam Tips

Common Pitfalls

Tips for Success

Practice & Assessment

Multiple Choice Questions

Q1.According to Kirchhoff's junction rule, if I1=3 AI_1 = 3 \, \text{A}I1​=3A and I2=2 AI_2 = 2 \, \text{A}I2​=2A enter a junction, and I3I_3I3​ leaves the junction, what is the value of I3I_3I3​?

Correct Answer: B

Q2.If the resistivity of a material is 1.5 x 10^-8 Ωm and the length and cross-sectional area of the conductor are 2 m and 1 x 10^-6 m² respectively, what is the resistance of the conductor?

Correct Answer: A

Q3.In a circuit, two cells with emfs of 9 V and 6 V are connected in series with internal resistances of 1 Ω and 2 Ω respectively. What is the total emf of the combination?

Correct Answer: A

Q4.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

Q5.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

Q6.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

Q7.Which of the following materials is likely to have a resistivity in the range of 10−8 Ω⋅m10^{-8} \, \Omega \cdot m10−8Ω⋅m to 10−6 Ω⋅m10^{-6} \, \Omega \cdot m10−6Ω⋅m?

Correct Answer: B

Correct Answer: C

Q9.Consider a circuit where a voltage source of 15 V is connected in series with three resistors of resistances 2 Ω, 3 Ω, and 5 Ω. What is the total current flowing through the circuit?

Correct Answer: A

Q10.Ohm's Law is given by V=IRV = IRV=IR. If a conductor has a resistance of 5 Ω and a current of 3 A flows through it, what is the potential difference across the conductor?

Correct Answer: A

Q11.In a Wheatstone bridge, four resistors of 10 Ω, 20 Ω, 30 Ω, and 40 Ω are connected. A galvanometer is connected across the bridge. If the bridge is balanced, what is the ratio of the resistances in the two arms?

Correct Answer: A

Q12.What happens to the current in a conductor if the electric field is reversed while keeping its magnitude constant?

Correct Answer: C

Q13.In a Wheatstone bridge, if the resistors are 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: B

Q14.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

Q15.In a circuit with nodes labeled A, B, C, and D, if the current entering node A is I and the current leaving node B is I/2, what is the current leaving node C assuming no other currents are involved?

Correct Answer: A

Q16.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

Q17.In a closed loop of a circuit, the potential difference across the loop is zero. If the loop contains three resistors with resistances R1R_1R1​, R2R_2R2​, and R3R_3R3​, and a voltage source VVV, what is the correct equation for the loop?

Correct Answer: A

Q18.Which of the following statements is true about the resistivity of materials?

Correct Answer: C

Q19.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

Q20.Which of the following statements about Kirchhoff's rules is correct?

Correct Answer: A

Q21.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

Q22.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

Q23.In a circuit, if the current III splits into two branches at a junction, with one branch carrying I1=I/3I_1 = I/3I1​=I/3, what is the current in the other branch?

Correct Answer: A

Q24.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

Q25.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

Q26.What is the equivalent resistance of two resistors, R1R_1R1​ and R2R_2R2​, connected in parallel?

Correct Answer: B