Topic practiceQ1Moderate
This question uses a figure from the original paper. The figure is being added; the answer and explanation are already verified.
Obtain the voltage v in the branch shown in Fig. Q6 for i_2 = −2 A.
View answer & solution
Correct answer: B. −20 V
Quick explanation
The controlling current is signed: i_2 = −2 A. Hence v_x = 15i_2 = −30 V. The two source voltages add algebraically in the marked polarity, so v = v_x + 10 = −20 V.
vx = 15i2
v = vx + 10
v = 15(−2) + 10 = −20 V
i2 = −2 A
vx = 15i2
vx = 15(−2) = −30 V
v = vx + 10 V
v = −30 + 10
v = −20 V
Bottom → top rise = 10 + (−30) = −20 V