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Answer to Question #6454 in Electric Circuits for Fainah
Question #6454
A 58.0 Ohms resistor is connected in parallel with a 124.0 Ohms resistor. This parallel group is connected in series with a 19.0 Ohms resistor. The total combination is connected across a 15.0 V battery.
(a) Find the current in the 124.0 Ohms resistor. ___________A
(b) Find the power dissipated in the 124.0 Ohms resistor.____________W
(a) Find the current in the 124.0 Ohms resistor. ___________A
(b) Find the power dissipated in the 124.0 Ohms resistor.____________W
Expert's answer
(a) Find the current in the 124.0 Ohms resistor. ___________A
Let's find the total resistance of the circuit:
R = 1/(1/58+1/124) + 19 = 58.5165 Ohms.
The total current is
I = U/R = 15/58.5165 = 0.2563 A.
It's the current through the 19.0 Ohms resistor and it is the same as a current through the 58.0 Ohms resistor and 124.0 Ohms resistor together.
The common resistance of paralleled 58.0 Ohms resistor and 124.0 Ohms resistor is
R12 = 1/(1/58+1/124) = 39.5165 Ohms.
U12 = I*R12 = 0.2563*39.5165 = 10.1281
So, the required current is
I1 = U12/124 = 10.1281/124 = 0.0817 A.
(b) Find the power dissipated in the 124.0 Ohms resistor.____________W
P = U12*I1 = 10.1281*0.0817 = 0.8272 W.
Let's find the total resistance of the circuit:
R = 1/(1/58+1/124) + 19 = 58.5165 Ohms.
The total current is
I = U/R = 15/58.5165 = 0.2563 A.
It's the current through the 19.0 Ohms resistor and it is the same as a current through the 58.0 Ohms resistor and 124.0 Ohms resistor together.
The common resistance of paralleled 58.0 Ohms resistor and 124.0 Ohms resistor is
R12 = 1/(1/58+1/124) = 39.5165 Ohms.
U12 = I*R12 = 0.2563*39.5165 = 10.1281
So, the required current is
I1 = U12/124 = 10.1281/124 = 0.0817 A.
(b) Find the power dissipated in the 124.0 Ohms resistor.____________W
P = U12*I1 = 10.1281*0.0817 = 0.8272 W.
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Thank you so much! I failed to register that part a was referring to a single resistor and not overall (The wording from my book did not state so clearly) Lifesaver!
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