
A current of 3 A flows through the $2 \Omega$ resistor shown in the circuit. The power dissipated in the $5…

- $4 \mathrm{~W}$
- $2 \mathrm{~W}$
- $1 \mathrm{~W}$
- $5 \mathrm{~W}$
Solution
$V=2 \times 3=6 \mathrm{~V}$
So, voltage across lowest arm,
$V_1=6 \mathrm{~V}$
Current across $5 \Omega, I=\frac{6}{1+5}=1 \mathrm{~A}$
Thus, power across $5 \Omega$,
$P=I^2 R=(1)^2 \times 5=5 \mathrm{~W}$
Asked in: NEET 2008 (Screening)