Discussion
The current \(i_{1}\) in the following network is given by
Answer: B
The current is \(i_{1} = -4.096\) mA. Take the bottom wire as ground, with kilo-ohms and milliamps giving volts. The dependent source is 4 ohm times \(i_{1}\), so it contributes only about \(0.004\,i_{1}\) volts, negligible beside the kilo-ohm drops. Node equations for the left node \(V_{A}\) and the right node \(V_{B}\):
\[ V_{A} = -47(1 + i_{1}) \]
\[ V_{B} = 10(i_{1} + 20) \]
\[ V_{B} = V_{A} - 3.3\,i_{1} \]
Substituting gives \(10 i_{1} + 200 = -47 - 47 i_{1} - 3.3 i_{1}\), so \(60.3\, i_{1} = -247\) and \(i_{1} \approx -4.096\) mA. The positive values in (C) and (D) have the wrong sign, and \(-40.96\) mA is ten times too large.
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