Question #321347

A 32.0-Ω resistor and a 20.0-Ω are connected in parallel, and the combination is applied



across a 240 V dc line.



(a) What is the resistance of the parallel combination?



(b) What is the total current through the parallel combination?



(c) What is the current through each resistor?




Expert's answer

a) Resistance of the parallel combination:

R=R1∗R2R1+R2=32.0∗20.032.0+20.0≈12.3(Ohm)R = \cfrac{R1*R2}{R1+R2} = \cfrac{32.0*20.0}{32.0+20.0} \approx 12.3 (Ohm)



b) Total current through the parallel combination

I=VR=24012.3≈19.5(A)I = \cfrac{V}{R} = \cfrac{240}{ 12.3} \approx 19.5 (A)



c) Current through each resistor

I1=VR1=24032.0=7.5(A)I2=VR2=24020.0=12(A)I1 = \cfrac{V}{R1} = \cfrac{240}{32.0} = 7.5 (A)\\ I2 = \cfrac{V}{R2} = \cfrac{240}{20.0} = 12 (A)



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