Answer to Question #243387 in Electrical Engineering for Neilmar

Question

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?

Answer 1

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:

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