In the network shown in the figure below, the current in the 12 Ohm resistor is 5 Amps, the battery voltage E is [Figure description: A single-loop-with-parallel-branches DC circuit. A battery E is on the left vertical side; a 4 Ω resistor is in series on the top wire. After the 4 Ω resistor the circuit splits into two parallel vertical branches connected between the top and bottom wires: (1) a middle branch containing an 18 Ω resistor; (2) a right branch containing a 6 Ω resistor in series with a parallel combination of a 4 Ω resistor and a 12 Ω resistor. The 5 A current is marked as flowing through the 12 Ω resistor.]
- A. 120 Volts
- B. 180 Volts
- C. 300 Volts
- D. 310 Volts
Answer: C. 300 Volts
Explanation: 12 Ω carries 5 A (60 V across 4∥12 = 3 Ω, so 15 A in the 4 Ω, 20 A in the 6 Ω branch). Branch voltage = 60 + 6×20 = 180 V; 18 Ω carries 10 A; total 30 A; E = 180 + 4×30 = 300 V.