In this post, we will solve *numerical questions or problems related to the emf of a cell and internal resistance*. We will also use the concepts of terminal potential difference, Open circuit, short circuit, & closed circuit, V-I curve, etc. to solve the problems using equations like V = E – Ir.

## emf and internal resistance questions – numerical problems solved

**Question 1) **

**What is the internal resistance of a battery if its emf is 6 V and the potential difference across its terminals is 5.8 V when a current of 0.5 A flows in the circuit when it is connected across a load?**

**Solution:**

emf = E = 6 V

Terminal Potential Difference = V = 5.8 V

Current I = 0.5 A

Say, **internal resistance of a battery** = r

We will use this equation: V = E – Ir

r = (E – V)/ I = (6-5.8)/0.5 ohm = 0.4 ohm

Answer: **internal resistance of a battery** = 0.4 ohm

**Question 2) **

### A 12.0 V battery has an internal resistance of 7.0 ohm.

a) What is the maximum current this battery could supply?

b) What is the potential difference across its terminals when it is supplying a current of 150.0 mA?

c) Draw a sketch graph to show how the terminal potential difference varies with the current supplied if the internal resistance remains constant. How could the internal resistance be obtained from the graph?

**Solution**:

Emf E = 12 V

internal resistance r = 7 ohm

**a) **

Maximum current = I_{max} = ?

Current is maximum when the external resistance R (load) = 0 (in case of short circuit)

Hence, drop against this load = V =IR= 0. This means, the Terminal Potential difference = V = 0

We will use this equation: V = E – Ir

Putting, terminal PD = V = 0,

=> 0 = E – Ir

I_{max} = E/r = 12/7 amp = 1.7 amp

[ **alternate solution**: current I = emf / (load resistance + internal resistance)

=> I = E/(R + r)

For maximum current R = 0 (short circuit)

Hence, I_{max} = E/r = 12/7 amp = 1.7 amp]

**b) **

when I = 150 ma = 150/1000 amp, we have to find out the value of terminal PD = V

V = E – I r = 12 – (150/1000) . 7 = 10.95 V

**c) **

To draw V-I curve: we will use the equation: V = E – Ir

Open circuit: When I = 0, V = E = 12 V (here)

Short circuit: When V =0, then, current I = E/r = 1.7 amp.

The V-I graph can be obtained using these 2 sets of data.