Question

38. A 12-ohm resistor is connected in parallel with a series combination of resistors of 8 and 16 ohms. If the drop across the 8-ohm resistor is 48 volts, determine the total impressed emf and the total current.

Answer 1

Voltage across R1 = VR1 = 48V

Also R1 and R2 are in series, therefore, current through R1 and R2 will be same.

Current through R1 = I = $\frac{VR1}{R1} = \frac{48}{8} = 6A$

Therefore voltage across R2 = VR2 = I * R2 = 6 * 16 = 96 V

Total voltage in branch = V = VR1 + VR2 = 48 + 96 = 144 V

As R3 is in parallel with series combination of R1 and R2. Hence voltage across R3 = V = 144v

Total impressed emf = 144 v

Total equivalent res. = R = R3 || (R1 + R2) = 12 || (8 + 16) = 12 || 24

R = $R = \frac{12 * 24}{12+24} = 8 Ohms$

Hence Total Current = $\frac{V}{R} = \frac{144}{8} = 18 A$