Question #313465

The circuit in the figure on the right shows a network of resistors


connected in series and in parallel.


(a) Determine the total resistance of the network.


(b) What is the current through the 3.00-Ω?

1
Expert's answer
2022-03-21T12:47:19-0400

Series combination


Rt=R1+R2+R3R_t=R_1+R_2+R_3

Rt=3+3+3=9ΩR_t=3+3+3=9\Omega

Parallel combination


1Rt=1R1+1R2+1R3\frac{1}{R_t}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}

1Rt=1R+1R+1R\frac{1}{R_t}=\frac{1}{R}+\frac{1}{R}+\frac{1}{R}

Rt=R3ΩR_t=\frac{R}{3}\Omega

(B)

R=3Ω\Omega Resistance current flow

Current

Assume (V)=3V

Series combination current

I=VRtI=\frac{V}{R_t}

I=39=0.33AI=\frac{3}{9}=0.33A

Parallel combination

Voltage drop all resistance equal

I=33=1AI=\frac{3}{3}=1A


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