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I've worked this problem, just want to confirm I worked it correctly.

**A---------R1-----------*-------------------**

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R2 R3

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B---------R4----------*--------------------

Req=5.25 Ohms, therefore total current, I, =V/Req 120V/2.25 Ohms I=22.86A

We then find the voltage drop over R1, since it sees all the current in the circuit

V=IR V=22.86A*1 Ohm v=22.86

we also know that R4 will see all current in the circuit, so we calculate its voltage drop as well.

v=68.58

the total of these two drops subtracted from the total equals 28.56.

Since R2 and R3 are in parallel, they see equal current, meaning the remaining voltage is dissipated over these two resistors.

using the power formula, P=V^2/R, we can find the power dissipated by R2 by 28.56^2/3.33 P=244.95 watts.

Is this correct, or have I missed something somewhere

| |

R2 R3

| |

B---------R4----------*--------------------

Req=5.25 Ohms, therefore total current, I, =V/Req 120V/2.25 Ohms I=22.86A

We then find the voltage drop over R1, since it sees all the current in the circuit

V=IR V=22.86A*1 Ohm v=22.86

we also know that R4 will see all current in the circuit, so we calculate its voltage drop as well.

v=68.58

the total of these two drops subtracted from the total equals 28.56.

Since R2 and R3 are in parallel, they see equal current, meaning the remaining voltage is dissipated over these two resistors.

using the power formula, P=V^2/R, we can find the power dissipated by R2 by 28.56^2/3.33 P=244.95 watts.

Is this correct, or have I missed something somewhere