Question

What is the power dissipated by each lamp of resistance $60\Omega$?

• A. 1 W
• B. 2 W
• C. 6 W
• D. 12 W

Answer 1

Answer: (C)

6 W

Resistance are in series

So total resistance $R_T=R+R=6+6=12$

current, $I=\frac{V}{R_T}=\frac{12}{12}=1A$

Power dissipated across each reistance $P=i^{2}R=1^2\times 6=6W$