Question

$A B$ and $C D$ are two uniform resistance wires of lengths $100 \, cm$ and $80 \, cm$, respectively. The connections are shown in the figure. The cell of emf $5\, V$ is ideal while the other cell of emf $E$ has an internal resistance of $2\, \Omega$. A length of $20\, cm$ of wire $C D$ is balanced by $40 \, cm$ of wire $A B$. Find the emf $E$ (in volt), if the reading of the ideal ammeter is $2\, A$. The other connecting wires have negligible resistance.

Answer 1

Potential difference across wire $A B=5\, V$
$\therefore $ Potential difference across $40\, cm$ of this wire
$=\frac{5}{100} \times 40=2\, V$
$\therefore $ Potential difference across $20\, cm$ of wire $C D$
$=2\, V$
$\therefore $ Potential difference across wire $C D$
$=\frac{2}{20} \times 80=8\, V$
Potential difference across $2 \, \Omega$ resistor
$=2 \times 2=4\, V$
$\therefore $ Emf of the cell $=12\, V$