Question

The emf of the battery $\epsilon$ in the circuit shown in figure is 15 volt and internal resistance is 0.5 ohm. What is the current drawn from the battery?

• A. 1 A
• B. 2 A
• C. 3 A
• D. 4 A

Answer 1

Answer: (A)

The resistances of $7\Omega ,1\Omega$ and $10\Omega$ are in series, their equivalent resistance is
$R_S=7\Omega +1\Omega +10\Omega =18\Omega$
The equivalent circuit is shown in figure (a).

Now, as $R_S(=18\Omega )$ and $6\Omega$ are in parallel and is equivalent to
$\frac{6\times 18}{6+18}=4.5\Omega$
The equivalent circuit diagram is shown in figure (b).
So the equivalent resistance of the circuit will be

$R_{eq}=0.5+2+4.5+8=15\Omega$
Current drawn from the battery is
$I=\frac{V}{R_{eq}}=\frac{15V}{15\Omega }=1A$