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A current of $0.1$ A flows through a $25 ~\Omega$ resistor represented by the circuit diagram. The current in the $80 ~\Omega$ resistor is
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Verified Answer
The correct answer is:
$0.3 \mathrm{~A}$
$0.1 \times\left(25+\frac{20 \times 60}{20+60}\right)=i_{2} \times 20$
$I_{2}=0.2 \mathrm{~A}$
Hence, $\mathrm{i}$ through $80 \Omega$
$0.1+0.2=0.3 \mathrm{~A}$
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