Question

(8) For the electric circuit shown i. Determine the voltage across the 5.0 Ω resistor. ii. Which end of this resistor is at the higher potential? 5.0 Ω 10.0 Ω Solution: 0.75 V, left end 10.0 Ω 10.0 V 5.0 V T2.0 V

Answer 1

i) consider ABCFA loop we get 10 + 5i_1 - 10i_2 - 2 = 0 5i_1 - 10i_2 = -8 ---(1) consider CDEFC loop we get At junction C i = i_1 + i_2 Apply kirchoff voltage law to CDEFC loop we get 10(i_1 + i_2) - 15 + 2 + 10i_2 = 0 10i_1 + 20i_2 = 13 -------(2) i_1 + 2i_2 = 1.3 ------(2) on solving the eq(1) and eq(2) we get i_1 = -0.15a i_2 = 0.725A voltage across 5 Ohm is V_1 = i_1(5 Ohm) = (-0.15)5 V_1 = -0.75V V_C + (0.15)5 = V_b V_C - V_b = -0.75v so v_b is at higher potential left end is at higher potential.