Figure shows a part of an electric circuit. The potentials at points a, b and c are 30 V, 12 V and 2 V respectively. The current through the 20 Ω resistor will be,

Solution

Physics sir · 4 steps

Call the junction potential V, balance current in and current out, and the whole circuit opens up.

1. Name the junction. Let the point where the three resistors meet be at potential V. Current flows from a into this junction through 10 Ω, and leaves through 20 Ω to b and through 30 Ω to c.
2. Apply the junction rule. Kirchhoff's current law says current in equals current out. Each current is the potential difference across that resistor divided by its resistance.

   (30 - V)/10 = (V - 12)/20 + (V - 2)/30

3. Solve for V. Multiply through by 60: 6(30 - V) = 3(V - 12) + 2(V - 2), so 180 - 6V = 5V - 40. That gives 11V = 220, so V = 20 V.

   Resistor	Potential difference	Current
   10 Ω (a to junction)	30 - 20 = 10 V	1.0 A
   20 Ω (junction to b)	20 - 12 = 8 V	0.4 A
   30 Ω (junction to c)	20 - 2 = 18 V	0.6 A

4. So. The current through the 20 Ω resistor is 8/20 = 0.4 A. As a check, 0.4 + 0.6 = 1.0 A, which matches the current coming in.

The idea to keep. When potentials at the ends are given, take the unknown junction potential as V and write current in = current out. One equation, one unknown, then every branch current follows from Ohm's law.

Answer from the NTA final answer key. Solution by PrepWiser.