- #1

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**1. The 10 ohm resistor in the figure is dissipating 40W of power. How much power is the 5ohm resistor dissipating?**

**2. P = I**

^{2}*R**3. I = sqrt(P/R) = 2. I then did P = I**

^{2}R = (4)(5) = 20 which is incorrect.-
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- Thread starter Linus Pauling
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- #1

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- #2

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So, we have 10 ohm and 40 watt.

the Power = V^2/ Ohm then we can calculate the volt on 10 ohm.

The power on 20 ohm then, is // resistor with 10 ohm.

Then they should have same potential difference.

So you can find the power at 20 ohm...

BUt......

I am lost with 5 ohm too... anybody else?

- #3

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Since P = I^2 R, we need to find I1 in order to calculate the power dissipated by the 5 ohm resistor. We use the loop rule to find current I3, since we already know I2 = 2 amps. So draw Loop 1 clockwise around the inside of the parallel part of the circuit (containing the 10 ohm and 20 ohm resistors). The sum of the voltage drops (V=IR) across the circuit equals zero. We start at the junction A, and determine the voltage drop across each resistor around the loop.

The Loop 1 equation gives I3 = (1/2)I2. Substituting I3 into the junction A equation, with I2 = 2 amps, we get I1 = (3/2)I2 = 3 amps. So the power dissipated by the 5 ohm resistor is P = I^2 R = (3 amps)^2 * (5 ohms) = 45 watts. Feel free to make any corrections.

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