Electric Current Practice Questions

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Electric current is a topic covered under the Current Electricity chapter in class 12 Physics syllabus. The important questions that are included in this chapter are covered in this article. Mainly numerical based questions are asked from this topic.

Electric current is defined as the rate of flow of electrons in a conductor. When an external force is applied, the negatively-charged electrons start moving from negative to positive terminal of a circuit. This flow of electrons generates electric current. The external force due to which electrons flow is called electromotive force. Electromotive force is measured in volts.

Current flows in the direction opposite to that of flow of electrons.

The conventional current flow is from the positive to negative terminal i.e. in the direction of flow of positive charges.
The movement of electrons is from negative to positive terminal in the circuit.

Check Also: NCERT Solutions for Class 12 Physics Current Electricity

Important Questions on Electric Current

Ques 1. Define the mobility of charge carriers in a conductor. Write its SI unit also. (2 Marks)

Ans. The magnitude of the drift velocity of charge carriers per unit applied electric field is called mobility of charge carriers. Its SI unit is m²V^-1s^-1.

Ques 2. A 10 V cell of negligible Internal resistance Is connected In parallel across a battery of emf 200 V and internal resistance 38Ω as shown in the figure. Find the value of current in the circuit. (2 Marks)

A 10 V cell of negligible Internal resistance Is connected In parallel across a battery of emf 200 V and internal resistance 38Ω as shown in the figure

Ans. Given E₁ = 10 V, r₁ = 0, E₂ = 200 V, r₂ = 38 Ω

The two cells send current in the opposite direction, therefore, net emf of the combination

E_net = 200 – 10 = 190 V

Hence, current in the circuit is

I = E_net / R

I = 190 / 38

I = 5 A

Ques 3. The circuit in the figure shows two cells connected in opposition to each other. Cell E₁ is of emf 6 V and internal resistance 2 Ω; the cell E₂ is of emf 4 V and internal resistance 8 Ω. Find the potential difference between the points A and B. (3 Marks)

The circuit in the figure shows two cells connected in opposition to each other.

Ans. The cells in the circuit shown above are connected in opposite directions, therefore the net EMF is;

E = E₁ - E₂

E = 6 - 4 = 2 V

Hence the current in the circuit is,

I = E / R + r

I = 2 / 10

I = 0.2 A

Potential difference across E₁ = 6 - 0.2 × 2 = 5.6 V

Potential difference across E₂ = VAB = 4 + 0.2 × 8 = 5.6 V

Ques 4. If an electric field of magnitude 570 NC^-1 is applied in the copper wire, find the acceleration experienced by the electron. (3 Marks)

Ans. E = 570 NC^-1, e = 1.6 × 10^-19 C, m = 9.11 × 10^-31 kg, a = ?

F = ma = eE

a = eE / m

a = 570 × 1.6 × 10^-19/ 9.11 × 10^-31

a = 912 × 10^-19 × 10³¹ / 9.11

a = 1.001 × 10¹⁴ ms^-2

Ques 5. A copper wire of cross-sectional area 0.5 mm² carries a current of 0.2 A. If the free electron density of copper is 8.4 × 10²⁸ m^-3 then compute the drift velocity of free electrons. (3 Marks)

Ans. The relation between the drift velocity and current in the wire is given by;

V_d= I / ne A

V_d= 0.2 / 8.4 × 10²⁸ × 1.6 × 10^-19 × 0.5 × 10^-6

V_d= 0.03 × 10^-3 ms^-1

Read Also:

Electric Current Related Topics
Types of Current	Electric Current and Circuit	Circuit Diagram
Kirchhoff’s Law	Combination of Resistors	Combination of Cells
Alternating Current	Difference Between Ammeter and Voltmeter	Unit of Current
Network Analysis	Resistance	Uses of Resistor

Ques 6. Determine the number of electrons flowing per second through a conductor when a current of 32 A flows through it. (3 Marks)

Ans. I = 32 A, t = 1 sec

Charge of an electron, e = 1.6 × 10^-19 C

Let n be the number of electrons flowing per second

I = Q / t

I = ne / t

n = It / e

n = 32 × 1 / 1.6 × 10^-19

n = 20 × 10¹⁹

n = 2 × 10²⁰ electrons

Ques 7. Calculate the equivalent resistance for the circuit which is connected to 24 V battery and also find the potential difference across 4 Ω and 6 Ω resistors connected in series in the circuit. (3 Marks)

Ans. The two resistors are connected in series, therefore the effective resistance in the circuit is

Req = 4 Ω + 6 Ω = 10 Ω

Current in the circuit is, I = V / R_eq= 24 / 10 = 2.4 A

Voltage across 4 Ω resistor,

V₁ = IR₁ = 2.4 × 4 = 9.6 V

V₂ = IR₂ = 2.4 × 6 = 14.4 V

Ques 8. Calculate the equivalent resistance in the following circuit and also find the current I, I₁ and I₂ in the given circuit. (5 Marks)

Calculate the equivalent resistance in the following circuit and also find the current I, I1 and I2 in the given circuit

Ans. The resistances are connected in parallel, so the net resistance is given by;

1/R_p = 1/R₁ + 1/R₂

1/R_p = 1/4 + 1/6

1/R_p = 5/12 Ω

R_p = 12/5 Ω

The resistors are connected in parallel and the voltage across each resistor is same, so

I₁ = V/R₁ = 24/4 = 6 A

I₂ = V/R₂ = 24/6 = 4 A

Total Current, I = I₁ + I₂ = 6 + 4 = 10 A

Ques 9. Find the value of I from the given circuit. (3 Marks)

Find the value of I from the given circuit

Ans. Applying Kirchhoff’s rule to the center point of the circuit. The arrows pointing towards the center are positive and those away from the center are negative.

So, 0.2 A - 0.4 A + 0.6 A - 0.5 A + 0.7 A - I = 0

1.5 A - 0.9 A - I = 0

0.6 A - I = 0

I = 0.6 A

Ques 10. Calculate the current that flows in the 1 Ω resistor in the following circuit. (5 Marks)

Calculate the current that flows in the 1 Ω resistor in the following circuit

Ans.

From the above figure, we can say that the current flowing through the 9 V battery as I₁ splits into I₂ and I₁ - I₂ (according to Kirchhoff’s current rule).

Consider the loop EFCBE, and apply KVR, we get

1I₂ + 3I₁ + 2I₁ = 9

5I₁ + I₂ = 9 ……..(1)

Applying KVR to the loop EADFE, we get

3 (I₁ - I₂) - 1I₂ = 6

3I₁ - 4I₂ = 6 ………(2)

Solving eq. (1) and (2), we get

I₁ = 1.83 A

I₂ = -0.13 A

The negative sign indicates that the current in the 1Ω resistor flows from F to E.

Check Out:

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Class 12 Physics Book	Class 12 Biology Book	Class 12 Maths Book
CBSE Class 12 Notes	Chemistry MCQs	Important Maths Formulas
Chemistry Study Guide	Biology Study Guide	Physics Study Guide
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Class 12 Biology Syllabus	Class 12 Maths Syllabus	Class 12 English Syllabus

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Electric Current Practice Questions

Important Questions on Electric Current

CBSE CLASS XII Related Questions

2.
Derive an expression for the capacitance of a parallel plate capacitor of plate area A and plate separation d with air present between the plates.

4.
Two metal spheres of radii $r_1$ and $r_2$ ($> r_1$) having charges $q_1$ and $q_2$ respectively kept in air, are brought in contact. Which of the following statements is not correct ?

Similar Physics Concepts

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Electric Current Practice Questions

Important Questions on Electric Current

CBSE CLASS XII Related Questions

2.Derive an expression for the capacitance of a parallel plate capacitor of plate area A and plate separation d with air present between the plates.

4.Two metal spheres of radii $r_1$ and $r_2$ ($> r_1$) having charges $q_1$ and $q_2$ respectively kept in air, are brought in contact. Which of the following statements is not correct ?

Similar Physics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

2.
Derive an expression for the capacitance of a parallel plate capacitor of plate area A and plate separation d with air present between the plates.

4.
Two metal spheres of radii $r_1$ and $r_2$ ($> r_1$) having charges $q_1$ and $q_2$ respectively kept in air, are brought in contact. Which of the following statements is not correct ?