P 4.5-6 Simplify the circuit shown in Figure P 4.5-6 by replacing series and parallel resistors by equivalent resistors. Next, analyze the simplified circuit by writing and solving mesh equations.

(a) Determine the power supplied by each source.

(b) Determine the power absorbed by the 30-Ω resistor.

Figure P 4.5-6

So we can join the 60 Ω and the 40 Ω resistors into a single 100 Ω

We have the 60 Ω resistor in parallel with 300 Ω which leaves an affective resistor

We have the 80 Ω in parallel with a 560 Ω resistor providing an effective resistor

So this effective 70 Ω resistor is in series with the 30 Ω and the 100 Ω resistor creating an effective resistance of 200 Ω So putting these all together we now have the following effective circuit

Write out the Mesh Equations

Solve by multiplying second equation by 5 and adding the two equations

Solving for i₁

(a) Determine the power supplied by each source.

12 V supply

8 V supply

Note Power convention means i₂ is going backwards through power supply as drawn, that is where the outside – sign comes from. But i₂ actually is negative so we are getting a positive power which implies power supply is providing power not absorbing power.

(b) Determine the power absorbed by the 30-Ω resistor.

The current through the 30 Ω resistor is i₁, so power dissipated or absorbed by 30 Ω resistor is found from

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This page last updated on January 31, 2019