Question

The emf of the battery $\varepsilon$ 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

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