Question

Two resistors of resistance $3\Omega$ and $6\Omega$ are in parallel and the combination is in series with a $4\Omega$ resistor. A potential difference of 90V is applied across the network. The potential difference across the $4\Omega$ and the current in the $3\Omega$resistor are respectively,

• A. 30V : 5A
• B. 30V : 10A
• C. 60V : 5A
• D. 60V : 10A

Answer 1

The correct option is D

60V : 10A

$R_{eff}=\frac{1}{\frac{1}{3}+\frac{1}{6}}+4=\frac{6}{3}+4=6\Omega$ $\therefore i=\frac{90}{6}=15A$
$\therefore$ p.d. across $4\Omega$ resistance
$=i\times R=15\times 4=60V$
$\therefore$ P.d across the parallel combination = 30V
$\therefore$ Current through $3\Omega resistance=\frac{30}{3}=10A$