When a current of 0.2 A is drawn from a battery, then potential difference between its terminals is 20 V and when a current of 2 A is drawn, then the potential difference drops to 16 V. The emf of battery will be :

Question

When a current of 0.2 A is drawn from a battery, then potential difference between its terminals is 20 V and when a current of 2 A is drawn, then the potential difference drops to 16 V. The emf of battery will be : (A) 20.4 eV (B) 24.4 eV (C) 19.8 eV (D) 16.1 eV

Tardigrade Admin · Accepted Answer

When current is drawn potential difference is less than emf.
The emf (E) is the characteristic of each cell and its value remains constant for the cell, while the potential difference (V) goes on, decreasing on taking more and more current (i) from the cell.
So,

V = E - i r

where r is internal resistance of cell.
Given,

V_{1} = 20 V, i_{2} = 2 A, i_{1} = 0.2 A

First case:

20 = E - 0.2 r

\Rightarrow E = 20 = 0.2 r

... (i)
Second case:

16 = E - 2 r

\Rightarrow E = 16 + 2 r

... (ii)
Solving Eqs. (i) and (ii),
we get

E = 20.4 V

Note: It is evident from the equation

V = E - i r

larger the current

i

drawn from the cell, smaller will be the potential difference across its plates.

Q. When a current of 0.2A is drawn from a battery, then potential difference between its terminals is 20V and when a current of 2A is drawn, then the potential difference drops to 16V. The emf of battery will be :

20.4 eV

24.4 eV

19.8 eV

16.1 eV

Q. When a current of $0.2 A$ is drawn from a battery, then potential difference between its terminals is $20 V$ and when a current of $2 A$ is drawn, then the potential difference drops to $16 V$ . The emf of battery will be :