Question

A part of the circuit is shown in figure. If B is earthened, find potential (in volts) of "F".

Answer 1

7.5 V

Answer 2

1. At node D, the voltage is 12 V. The resistor from D to E (2 Ω) has a voltage drop of 12 V – 5 V = 7 V.

   • Current from D to E = 7 V / 2 Ω = 3.5 A.

2. The resistor from C to D carries 5 A (into node D). By Kirchhoff’s Current Law at node D, the current that flows from D to F is:

   • $I_{DF}$ = 5 A – 3.5 A = 1.5 A.

3. The resistor from D to F is 3 Ω. The voltage drop across it equals:

   • $V_{drop}$ = 1.5 A × 3 Ω = 4.5 V.

4. Therefore, the potential at node F is:

   • $V_F$ = $V_D$ – $V_{drop}$ = 12 V – 4.5 V = 7.5 V.

Explanation (Minimal):

• Use KCL at node D: incoming current = 5 A; current to E = (12 – 5)/2 = 3.5 A; so, current towards F = 1.5 A.
• Voltage drop from D to F = 1.5×3 = 4.5 V.
• Hence, potential at F = 12 – 4.5 = 7.5 V.