Question

The emf of a cell E is $15V$ as shown in the figure with an internal resistance of $0.5\Omega$. Then the value of the current drawn from the cell is:

• A. $2A$
• B. $5A$
• C. $1A$
• D. $3A$

Answer 1

The correct option is C $1A$

$7\Omega ,1\Omega ,10\Omega$ resistances are connected in series.

Hence an equivalent resistance of $7+1+10=18\Omega$ is connected in parallel with a $6\Omega$ resistance.

Hence, equivalent resistance of $\frac{6\times 18}{6+18}\Omega =4.5\Omega$ is connected in series with $0.5\Omega ,8\Omega ,2\Omega$.

Hence, net resistance of the circuit is $4.5+0.5+8+2=15\Omega$

Hence, current drawn from cell=$\frac{E}{R_{eq}}=\frac{15V}{15\Omega }=1A$