In the circuit shown in the figure, R1 = 3.00 Ω, R2 = 6.00 Ω, R3 = 20.0 Ω, and Vemf = 12.0 V. will give fave ans physics?

Question

In the circuit shown in the figure, R1 = 3.00 Ω, R2 = 6.00 Ω, R3 = 20.0 Ω, and Vemf = 12.0 V.

a. Determine a value for the equivalent resistance.

b. Calculate the magnitude of the current flowing through R3 on the top branch of the circuit (marked with a vertical arrow).

NCS · Accepted Answer

This is made a heck of a lot easier due to the symmetry of the top and bottom. For the R1/R2 in parallel,
R12 = 1/(1/3.00 + 1/6.00) = 1/(3/6.00) = 2.00 Ω
and so for the top and bottom halves of the circuit
Rt = Rb = 2*2.00Ω + 20.0Ω = 24.0Ω

Applying Kirchoff's Law on the top loop:
E = 12.0 V = It * Rt = It * 24.0Ω → → It = 0.500 A ◄ (b)
and of course for the bottom we get Ib = 0.500 A

(a) Not sure about this part. I'd wager that
Req = 12.0V / 0.500A = 24.0 Ω
but it's possible that we should use the total current, and so
Req = 12.0V / 2*0.500A = 12.0 Ω

Philomel · Answer

b.
  3//6=2Ω
2+20+2+=24Ω
Rt=24//24=12Ω

a.
It=12/12=1A
I(3Ω)=1/2A

In the circuit shown in the figure, R1 = 3.00 Ω, R2 = 6.00 Ω, R3 = 20.0 Ω, and Vemf = 12.0 V. will give fave ans physics?

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