Question:
In the figure shown, what is the current (in Ampere) drawn from the battery? You are given :
$\mathrm{R}_{1}=15 \Omega, \mathrm{R}_{2}=10 \Omega, \mathrm{R}_{3}=20 \Omega, \mathrm{R}_{4}=5 \Omega, \mathrm{R}_{5}=25 \Omega, \mathrm{R}_{6}$
$=30 \Omega, \mathrm{E}=15 \mathrm{~V}$
Correct Option: , 3
Solution:
(3) $R_{3}, R_{4}$ and $R_{5}$ are in series so their equivalent $R=20+5+25=50 \Omega$
This is parallel with $R_{2}$, and so net resistance of the circuit
$R_{\mathrm{eq}}=\left(\frac{10 \times 50}{10+50}\right)+15+30=\frac{160}{3} \Omega$
So, $i=\frac{\varepsilon}{R_{\mathrm{eq}}}=\frac{15}{(100 / 3)}=\frac{9}{32} A$
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