help needed with physics problem

Question

20Ω

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5Ω 10Ω

The 10Ω resistor in is dissipating 55 W of power.

How much power is the 5Ω resistor dissipating?

How much power is the 20Ω resistor dissipating?

Answer 1

The key observation is that the 5Ω resistor has the same current as the 10Ω,

and the 20Ω resistor has the same voltage as the two resistors on the lower branch.

1) The 5Ω resistor:

By Ohm's law the voltage drop across the 5Ω resistor is half the voltage drop

accross the 10Ω resistor.

Since the current is the same, the power is half.

Therefore the power dissipating by the 5Ω resistor is 27.5W.

2) The 20Ω resistor:

The voltage drop across the 20Ω resistor is the same as the voltage drop

over the 15Ω of resistance on the lower branch.

By Ohm's law the current through the 20Ω resistor is smaller than

the current through the 15Ω by a factor of 15/20.

Thus the same factor is applied to the power.

The total power dissipated by the lower branch is 82.5W, therefore the

power dissipated by the 20Ω resistor is

82.5×15/20 = 61.875W

Answer 2

55 W = I²R. R = 10 Ohms, so I² = 5.5 A². Thus I = 2.32 A. Note that I is just the current through the 10 Ohm resistor. The 5 Ohm is in series with the 10 Ohm, so they share the same current. Hence, 5 Ohm power is (2.32)^2*5 = 5.5*5 = 27.5 W.

The 20 Ohm is in parallel with the combination of the 5 and 10 Ohm resistors, so the 20 Ohm voltage drop is the same as the sum of the voltage drops of the 5 and 10 Ohm resistors. Vdrop = Vdrop_5Ohm + Vdrop_10Ohms = I*(5 Ohms + 10 Ohms) = 2.32*15 = 34.8V.

Power dissipated in 20 Ohm resistor is thus P = I*V_20Ohms = (V_20Ohms)^2/R = (34.8)²/(20), or around 60.5 W