﻿ NEET Important Questions of Current Electricity | Zigya

## Previous Year Papers

Download Solved Question Papers Free for Offline Practice and view Solutions Online.

## Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

## Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.

# Current Electricity

#### Multiple Choice Questions

1.

An inductor (L = 0.03 H) and a resistor (R = 0.15 kΩ) are connected in series to a battery of 15V EMF in a circuit shown. The key K1 has been kept closed for a long time. Then at t = 0, K1 is opened and key K2 is closed simultaneously. At t = 1 ms, the current in the circuit will be:
()

• 100 mA

• 67 mA

• 6.7 mA

• 0.67 mA

D.

0.67 mA

After long time inductor behaves as short-circuit.
At t = 0, the inductor behaves as short circuit.The current,

As K2 is closed, current through the indicator starts decay which is given at any time t as

486 Views

2.

In the circuit shown, the current in the 1Ω resistor is:

• 1.3 A from P to Q

• 0 A

• 0.13 A, from Q to P

• 0.13 A, from P to Q

C.

0.13 A, from Q to P

Connect Point Q to ground and by applying Kirchhoff's laws
consider the grounded circuit as shown below,

Applying Kirchhoff's law at point Q,
Incoming current at Q = outgoing current from Q

Thus, current in the 1Ω resistance is 0.13 A, from Q to P

205 Views

3.

The coercivity of a small magnet where the ferromagnet gets demagnetized is 3 x 103Am-1.The current required to be passed in a solenoid of length 10 cm and number of turns 100, so that the magnet gets demagnetized when inside the solenoid, is

• 30 mA

• 60 mA

• 3 A

• 6 A

C.

3 A

For solenoid, the magnetic field needed to be magnetised the magnet,
B =μonl
Where, n = 100
l = 10 cm = 10/100 m  = 0.1m

330 Views

4.

The current voltage relation of the diode is given by I = e(1000V/T-1) mA, where the applied voltage V is in volt and the temperature T is in kelvin. If a student makes an error measuring ± 0.01V while measuring the current of 5mA at 300K, what will be the error in the value of current in mA?

• 0.2 mA

• 0.02 mA

• 05 mA

• 0.05 mA

A.

0.2 mA

Given, I =(e1000V/T-1)mA
dv =±0.01 V
T = 300 K
I = 5 mA
I = e1000V/T-1
I +1 = e1000V/T
Taking log on both sides, we get
log(l+T) = 1000V/T
On differentiating,

So, error in the value of current is 0.2 mA

283 Views

5.

When 5V potential difference is applied across a wire of length 0.1 m, the drift speed of electrons is 2.5 × 10–4 ms–1. If the electron density in the wire is 8 × 1028 m–3, the resistivity of the material is close to:

• 1.6 x 10-8Ωm

• 1.6 x 10-7Ωm

• 1.6 x 10-6Ωm

• 1.6 x 10-5Ωm

D.

1.6 x 10-5Ωm

vd = 25 x 10-4 m/s ⇒ n = 8 x 1028 /m3
we know that
J= nevd or I = nevd A
where, symbols have their usual meaning
331 Views

6.

The magnetic field in a travelling electromagnetic wave has a peak value of 20 nT. The peak value of electric field strength is :

• 3 V/m

• 6 V/m

• 9 V/m

• 12 V/m

B.

6 V/m

Ε = cB
Where E = electric field,
B = magnetic field
c = speed of EM waves
= 3 × 108 × 20 × 10–9 = 6 V/m

216 Views

7.

In the circuit shown here, the point C is kept connected to point A till the current flowing through the circuit becomes constant. Afterwards, suddenly point C is disconnected from point A and connected to point B at time t= 0. Ratio of the voltage across resistance and the inductor at t = L/R will be equal to

• 1

• -1

C.

-1

587 Views

8.

The supply voltage to a room is 120 V. The resistance of the lead wires is 6 Ω. A 60 W bulb is already switched on. What is the decrease of voltage across the bulb, when a 240 W heater is switched on in parallel to the bulb?

• zero

• 2.9V

• 13.3 V

• 10.04 V

D.

10.04 V

Resistance of bulb =(120x 120)/60 = 240Ω
Resistance of Heater =(120x 120)/240 = 60Ω

Voltage across bulb before heater is switched on,
Voltage across bulb after heater is switched on,
A decrease in the voltage is V1 − V2 = 10.04 (approximately)

348 Views

9.

This question has statement I and statement II. Of the four choices given after the statements, choose the one that best describes the two statements.
Statement- I: Higher the range, greater is the resistance of ammeter.
Statement- II: To increase the range of ammeter, additional shunt needs to be used across it.

• Statement – I is true, Statement – II is true, Statement – II is the correct explanation of statement- I.

• Statement – I is true, Statement – II is true, Statement – II is not the correct explanation of Statement–I.

• Statement – I is true, statement – II is false.

• Statement – I is false, Statement – II is true

D.

Statement – I is false, Statement – II is true

For Ammeter, S =
So for I to increase, S should decrease, so additional S can be connected across it.

234 Views

10.

Two long current carrying thin wires, both with current I, are held by insulating threads of length L and are in equilibrium as shown in the figure, with threads making an angle ‘θ’ with the vertical. If wires have mass λ per unit length then the value of I is: (g = gravitational acceleration)

B.

consider free body diagram of the wire
As the wires are in equilibrium. they must carry current in opposite direction.

Here, , where l is the length of each wire are d is a separation between wires.
From the diagram, d = 2L sin θ
T = cos θ = mg = λlg
(in vertical direction) ..... (i)

(In horizontal direction)  ..... (ii)
From eqs. (i) and (ii)

608 Views