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,

Step-by-step solution

Let the potential be x   V at the junction. By using V = I R we get, 30 - x = 10 i 1 x - 12 = 20 i 2 x - 2 = 30 i 3 By using Kirchhoff's law, i 1 = i 2 + i 3 ⇒ 30 - x 10 + 12 - x 20 + 2 - x 30 = 0 ⇒ 90 - 3 x + 18 - 1 . 5 x + 2 - x 30 = 0 ⇒ x = 20   V The current through the 20   Ω resistor is i 2 = 30 - 12 20   A ⇒ i 2 = 8 20 A = 0 . 4   A