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The 10 Ω resistor in FIGURE EX28.29 is dissipating 40Wof power. How much power are the other two resistors dissipating?

Short Answer

Expert verified

The power of other two resistors areP5Ω=20WandP20Ω=45W.

Step by step solution

01

Given Information 

We need to find that How much power are the other two resistors dissipating?

02

Simplify   

40Wof the power is dissipated by 10Ωresistor. So, calculating the current through the resistance 10Ω. The energy is dissipated when the current flows through a resistor. The rate of the dissipated energy is the power. This rate where the energy is transferred from the current to the resistor is

PR=I2R (1)

Use equation (1) and solve it for the current and plug the value for R=10Ωand PR=40Wto get I10Ωby

I10Ω=PRR=40W10Ω=2A

The current is the same for the resistors, and as the two resistors 10Ωand 5Ωare in series, the current I10Ωand I5Ωis same.

I10Ω=I5Ω=2A

Putting the values for the current I5Ω=2Aand R=5Ωinto equation (1) to get the dissipated power by resistance 5Ωby

P5Ω=I25ΩR=(2A)2(5Ω)=20W

The potential difference in the upper branch could be calculated by

V=I(10Ω+5Ω)=(2A)(10Ω+5Ω)=30V

The potential difference across the resistors is the same in parallel connection. So, the voltage across the below resistor is 30V. So, this value to get the power through the resistor 20Ωby

P20Ω=VR2R=30V20Ω=45W

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