Answer:
(a) 40.5 mA
(b) 5.06 W
(c) 0.473 W
Step-by-step explanation:
(a) The 12 V source and two 20 Ω resistors can be replaced by their Thevenin equivalent: 6 V in series with 10 Ω. Then that source and the 8 Ω resistor can be replaced by a Norton equivalent source of (6V)/(18Ω) = 1/3A in parallel with 18 Ω.
Then the current through the 40Ω resistor is proportional to its conductance in relation to the total conductance of 18Ω, 40Ω, 40Ω, and 10Ω, all in parallel. That is, the current is ...
(1/3A)((1/40)/(1/18 +1/40 +1/40 +1/10)) = (1/3A)(1/40)/(37/180) = 3/74 A
The current through either 40 Ω resistor is 3/74 A ≈ 40.5 mA
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(b) The equivalent resistance of all resistors to the right of the 8Ω resistor is ...
1/(1/40 +1/40 +1/10) = 6 2/3Ω
Then the load seen by the 12V source is 20Ω+(20Ω║(8Ω +6 2/3Ω)) = 28 6/13Ω. The power provided by the voltage source is then ...
P = V²/Req = (12V)²/(370/13Ω) = 5 11/185 W ≈ 5.059 W
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(c) Going back to the Thevenin equivalent voltage source, the 8Ω resistor is effectively in a series circuit with a total resistance of 10Ω +8Ω +6 2/3Ω = 24 2/3Ω and a source voltage of 6 V. Its power will be the square of the current, multiplied by its resistance.
P = I²·R = (6v/24 2/3Ω)²·8Ω = 648/1369 W ≈ 0.473 W