Discussion
The value of current (I) flowing in the circuit shown will be :
Answer: C
The current has a magnitude of 1.71 A.
Go round the loop in the direction of I. The source gives a rise of 24 V, the resistors drop \(4I\) and \(2I\), and since I enters the dependent source at its negative terminal, that source gives a further rise of \(5V_{1}\). With \(V_{1} = 4I\):
\[ \begin{aligned} 24 + 5V_{1} &= 4I + 2I \\ 24 + 20I &= 6I \\ I &= -\frac{24}{14} \approx -1.71\ \text{A} \end{aligned} \]
The minus sign means the current actually flows against the marked arrow; its size is 1.71 A, which is the value the options list. If the dependent source were taken as opposing the current, the loop would give \(24 = 26I\), or 0.92 A, which is not an option, so the polarity matters.
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