Question

In the figure shown, what is the current (in $\text{Ampere}$) drawn from the battery? You are given:
$R_1=15\Omega ,R_2=10\Omega ,R_3=20\Omega ,R_4=5\Omega ,R_5=25\Omega ,R_6=30\Omega ,E=15\text{V}$

• A. $20/3$
• B. $7/18$
• C. $13/24$
• D. $9/32$

Answer 1

The correct option is D $9/32$

$R_3,R_4$ and $R_5$ are in series, so their equivalent will be,

$R=20+5+25=50\Omega$

This is parallel with $R_2$ and so the net resistance of the circuit will be

$R_{\text{eq}}=15+\left(\frac{10\times 50}{10+50}\right)+30=\frac{160}{3}\Omega$

And, $I=\frac{E}{R_{\text{eq}}}=\frac{15}{\left(160/3\right)}=\frac{9}{32}\text{A}$

Hence, $\left(C\right)$ is the correct answer.