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 4$\Omega$ the resistor and the current in the 3$\Omega$ resistor are respectively,

• A.
• B.
• C.
• D.

Answer 1

The correct option is D

$R_{eff}=\frac{1}{\frac{1}{3}+\frac{1}{6}}+4=\frac{6}{3}+4=6\Omega \therefore i=\frac{90}{6}=15A$

$\therefore$ p.d. a cross $4\Omega$ resistance

$=i\times R=15\times 4=60V$

$\therefore$ P.d a cross the parallel combination $=30V$

$\therefore$ Current through $3\Omega$ resistance $=\frac{30}{3}=10A$