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 = I2R2 = 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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