Consider the following diagram, where V 1 = 14.5 V, V 2 = 10.5 V, R 1 = 86.0 m,

Question

Consider the following diagram, where V1 = 14.5 V, V2 = 10.5 V, R1 = 86.0 m, and R2 = 11.0 m.

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 defroster, and the like in parallel. If R = 0.800 , find the current in each branch. (Take the positive direction to be up.)

I alternator = ? (in A)
I battery = ? (in A)
I Resistor = ? (in A)

What is the terminal voltage of the battery?

Explanation / Answer

let the current in alternator is I1 (upwards)

current in battery is I2 (upwards) ,

apply KVL in alternate loop ,

14.5 - 0.086 * I1 - 0.80*(I1 + I2) = 0

0.886*I1 + 0.80*I2 = 14.5 .......eq1

aplly KVL in battery loop ,

10.5 - 0.011*I2 - 0.80*(I1 + I2) = 0

0.80*I1 + 0.811*I2 = 10.5 .......eq2

by solving eq1 and eq2,

I2 = -29.23 A

apply I2 in eq 1,

0.886*I1 - 0.80*29.23 = 14.5

I1 = 42.75 A

I alternator = 42.75 A

I battery = 29.23 A

I resistor = 42.75 - 29.23 = 13.52 A

(b)

terminal voltage of the battery = 10.5 - 29.23*0.011

Vb = 10.17 V

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