Question

In the circuit shown, the variable resistance is so adjusted that the ammeter reading is the same in both the positions 1 and 2 of the key. The reading of the ammeter is $2\text{A}$. If $E=10\text{V}$, then $X$ is

• A. $2\Omega$
• B. $5\Omega$
• C. $10\Omega$
• D. $20\Omega$

Answer 1

The correct option is B $5\Omega$

Ideal ammeter has no internal resistance.

Let $E_1$ and $r_1$ be the emf and internal resistance of the first cell.

When key is at (1) -
$E_1-I{r}_1=0$
$E_1=I{r}_1$
or $E_1=2r_1$

When key is at (2) -
$E_1-I{r}_1+E-IX=0$
$\Rightarrow 2r_1-2r_1+10-2X=0$
$\Rightarrow 10=2X$ or $X=5\Omega$