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Cells, EMF & Resistance (NCERT) - Free MCQ Practice Test with solutions,


MCQ Practice Test & Solutions: Test: Cells, EMF & Resistance (NCERT) (5 Questions)

You can prepare effectively for NEET NCERT Based Tests for NEET with this dedicated MCQ Practice Test (available with solutions) on the important topic of "Test: Cells, EMF & Resistance (NCERT)". These 5 questions have been designed by the experts with the latest curriculum of NEET 2026, to help you master the concept.

Test Highlights:

  • - Format: Multiple Choice Questions (MCQ)
  • - Duration: 5 minutes
  • - Number of Questions: 5

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Test: Cells, EMF & Resistance (NCERT) - Question 1

A cell having an emf ε and internal resistance r is connected across a variable external resistance R. As the resistance R is increased, the plot of potential difference V across R is given by

Detailed Solution: Question 1

image
Current in the circuit,

Potential difference across R,


When R = 0,V = 0
R = ∞,V = ε

Test: Cells, EMF & Resistance (NCERT) - Question 2

A battery of emf 15V and internal resistance of 4Ω is connected to a resistor. If the current in the circuit is 2A and the circuit is closed. Resistance of the resistor and terminal voltage of the battery will be

Detailed Solution: Question 2

Given, ε = 15V,
r = 4Ω,
I = 2A
Now, for resistance of the resistors
ε − Ir = V = IR;
15 − 2 × 4=2 × R;
15 − 8  = 2R
R = 7/2 = 3.5Ω.
Terminal voltage of battery,
V = IR = 2 × 3.5 = 7V

Test: Cells, EMF & Resistance (NCERT) - Question 3

The battery of a trunk has an emf of 24V. If the internal resistance of the battery is 0.8Ω. What is the maximum current that can be drawn from the battery?

Detailed Solution: Question 3

Here, ε = 24V and r = 0.8Ω
For the maximum current from the battery.
ε = Ir   (∵ R = 0)
∴ I = ε / r = 24/0.8 = 30A

Test: Cells, EMF & Resistance (NCERT) - Question 4

A battery having 12V emf and internal resistance 3Ω is connected to a resistor. If the current in the circuit is 1A, then the resistance of resistor and lost voltage of the battery when circuit is closed will be

Detailed Solution: Question 4

Here,
ε = 12V,  r = 3Ω, I = 1A, V = IR = ε − Ir
∴ 
= 12 − 3 = 9Ω
and V = IR = 1 × 9 = 9V

Test: Cells, EMF & Resistance (NCERT) - Question 5

When a current of 2A flows in a battery from negative to positive terminal, the potential difference across it is 12V. If a current of 3A flowing in the opposite direction produces a potential difference of 15V, the emf of the battery is

Detailed Solution: Question 5

Let ε be emf and r be internal resistance of the battery.
In first case,
12 = ε − 2r ...(i)
In second case,
15 = ε + 3r ...(ii)
Subtract (i) from (ii), we get
r  = 3 / 5Ω
Putting this value of r in eqn. (i), we get


= 66/5 = 13.2V

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