Physics

Problem

42. Three resistors

-R_{1}, R_{2}

, and

R_{3}

- are connected in series to a battery. Suppose

R_{1}

carries a current of

2.0 \mathrm{~A}, R_{2}

has a resistance of

3.0 \,\Omega

, and

R_{3}

dissipates 6.0 W of power. What is the voltage across

R_{3}

? (A) 1.0 V (B) 3.0 V (C) 6.0 V (D) 12 V (A) 1.0 V (B) 3.0 V (C) 6.0 V (D) 12 V

Intuitive explanation

In a series circuit, the same current flows through every component—so if $R_1$ carries 2.0 A, that's the current everywhere. The key insight is using the power dissipation formula $P = I^2 R$ (or equivalently $P = IV$ ) to find the voltage across $R_3$ . Since you know both the power dissipated and the current, you can find either the resistance or voltage directly.

Intuitive explanation

Browse the problem library