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

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.

## Comments

Assignment Expert04.03.13, 17:15You're welcome. We are glad to be helpful.

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Bonnie02.03.13, 22:22Thank 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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