Consider the following diagram, where V 1 = 13.5 V, V 2 = 9.5 V, R 1 = 78.0 m, a

Question

Consider the following diagram, where V1 = 13.5 V, V2 = 9.5 V, R1 = 78.0 m, and R2 = 10.0 m.

Consider the following diagram, where V 13.5 V, V2 9.5 V, R1 78.0 mn, and R2 10.0 mn. Altemator Battery The figure shows a simplified circuit diagram for an automobile. The equivalent resistor R represents the total electrical load due to spark plugs, lights, radio, fans, starter, rear window def and the like in parallel. If R 0.860 n, find the current in each branch. NTake the positive direction to be up.) Alternator F Battery IResistor What is the terminal voltage of the battery? Is the battery charging or discharging? charging discharging

Explanation / Answer

Let
1. I1 be the current coming out of positive pole of battery in branch containing R1
2. I2 be the current coming out of positive pole of battery in branch containing R2
3. let I3 be the current entering the load R
We have
V1 - I1*R1 = I3*R = V2 - I2*R2, and
I3 = I1 +I2 or
I1(R+R1) + I2* R = V1 ------------------- 1 and
I1*R + I2*(R+R2) = V2 --------------------- 2
1 -2 gives
I1*R1 - I2*R2 = V1 - V2 -------------------- 3
Adding 1*R2 and 3*R, we get
I1*{R2*(R+R1)+R1*R} = {(R2+R)V1 - R*V2} or
I1 = {(R2+R)V1 - R*V2}/{R2*(R+R1)+R1*R} or
I1 = {(0.01+0.86)*13.5 -0.86*9.5}/{0.01*(0.86+0.078)+0.078*0.86} or
I1 = 46.75 A
From 3 we have I2 = [{I1*R1-(V1-V2)}/R2] or
I2 = [46.75*0.078 - (13.5-9.5)]/0.01 = -35.35 A that is 35.35 into the battery Of voltage 12 V
and I3 = 46.75 - 35.35 = 11.4 A

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