**Suppose that we have a series circuit with a supply of 150 V and three resistors: R1 = 330 ohms, R2 = 680 ohms, and R3 = 910 ohms. What is the power dissipated by R2?**

SOLUTION

Find the current in the circuit. To do this, calculate the total resistance first.

Because the resistors are in series, the total is resistance is R = 330 + 680

+ 910 = 1920 ohms.

Therefore, the current is I = 150/1920 = 0.07813 A = 78.1 mA.

The power dissipated by R2 is P2 = I^{2}R2 = 0.07813 x 0.07813 x 680 = 4.151 W

We must round this off to three significant figures, getting **4.15 W.**

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