Question

A galvanometer together with an unknown resistance in series is connected across two identical batteries each of 1.5 V. When the batteries are connected in series, the galvanometer records a current of 1 A, and when the batteries are in parallel, the current is 0.6 A. What is the internal resistance of the battery?

• A. $1\Omega$
• B. $\frac{1}{2}\Omega$
• C. $\frac{3}{2}\Omega$
• D. $\frac{1}{3}\Omega$

Answer 1

The correct option is D

$\frac{1}{3}\Omega$

Let the internal resistance of each battery be r. Let R be the unknown resistance and G be the resistance of the galvanometer. Let E be the emf of each battery. When the batteries are connected in series, the total emf $=2E=2\times 1.5=3V$ and total internal resistance is 2r. The current in the circuit will be

$I=\frac{3}{R+G+2r}$

Given I= 1 A . Therefore

$1=\frac{3}{R+G+2r}orR+G=(3-2r)$ (1)

When the batteries are connected in paralle., the total emf = E = 1.5 V and the total internal resistance is r/2.

Hence the current in the circuit will be

$I^{,}=\frac{1.5}{R+G+\frac{r}{2}}orR+G=(2.5-\frac{r}{2})$ (2)

From Eqs. (1) and (2), we have

$3-2r=2.5-\frac{r}{2}$

Which gives $r=\frac{1}{3}$ohm.