Answer to Question #244957 in Electric Circuits for Ian

Question

. In the circuit shown in the figure, the ideal ammeter measures I = 0.4A in the indicated sense, and the ideal voltmeter measures a potential drop of V = 8.8V passing from b a a. Determines the value of the emf . Data: R = 56.212, R = 23.312, R = 27.4.2. & (A E = ? R: SR OR a

Answer 1

$R_1= 56.2 \;ohm \\ R_2 = 23.3 \; ohm \\ R_3 = 27.4 \;ohm \\ ε_2-R_2(I-I_1) -R_3I = 0 \\ V = I_1R_1 \\ I_1 = \frac{8.8 }{56.2} = 0.156 \;A \\ ε_2 - 23.3(0.4-0.156) -27.4(0.4) = 0 \\ ε_2 = 5.685 + 10.96 =16.645 \;V$