Answer to Question #243387 in Electrical Engineering for Neilmar

Question #243387
A 12Ω resistor is connected in parallel with a series combination of 8Ω & 16 Ω. If the drop across the 8
Ω resistor is 48 volts, Determine the total current?
1
Expert's answer
2021-09-30T02:39:57-0400




In above diagram,

the voltage drop across res. R2 = V2 = 48V

Therefore, current through res. R2 is given by:

I = V/R2 = 48/8 = 6A


Since R2 and R3 are in series, therefore, current is same through both res. R2 and R3.

Therefore, voltage drop across res. R3 = V3 = I * R3 = 6 * 16 = 96 V


Now the total voltage across the series combination of Res. R2 and R3 = V2 + V3 = 48 + 96 = 144 V.


Also the rse. R1 (=12 Oms) and series combination of (R2 and R3) are in parallel combination.

Hence, the equivalent resistance R = R1 || (R2 + R3)


\frac{1}{R} = \frac{1}{12} + \frac{1}{8+16}

On solving above, we get

R = 8 Ohms


The total current through the circuit is given by:


I = \frac{V}{R} = \frac{144}{8} = 18 Amp.


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