Refer to the four circuits shown.
Which one of the following options for \(k_1\), \(k_2\), \(k_3\), and \(k_4\) makes all of them realizable?
Show Hint
Look at which pairs of sources sit between the same two nodes: same-type sources, V-V or I-I, in parallel must match, while mixed V-I pairs never conflict.
Step 1: Look at how each pair of elements is joined.
In every one of the four circuits, the top wire and the bottom wire are plain wires with no components on them. So the left element and the right element both connect the same top node to the same bottom node. This means the two elements in each circuit are in parallel, not in series.
Step 2: Recall what parallel connection forces.
Two voltage sources connected in parallel between the same two nodes must have the same voltage, otherwise Kirchhoff's voltage law is broken. Two current sources connected in parallel between the same two nodes, feeding the node from the same side, must add up to zero at that node, otherwise Kirchhoff's current law is broken. A voltage source and a current source in parallel never conflict, because the voltage source can push or pull whatever current the current source needs, and the current source does not care what voltage appears across it.
Step 3: Analyze circuit 1, \(V_D\) parallel with \(V_X=k_1V_D\).
Both sources have the \(+\) mark on top, so they are in parallel with matching polarity. For this to hold for any nonzero \(V_D\),
\[
V_D=V_X=k_1V_D
\]
which gives
\[
k_1=1.
\]
Step 4: Analyze circuit 2, \(V_D\) parallel with \(I_X=0.5k_2V_D\).
This is a voltage source in parallel with a current source. Whatever current the current source demands, the voltage source simply supplies or absorbs it. There is no equation linking \(V_D\) and \(I_X\), so this circuit is realizable for every value of \(k_2\).
Step 5: Analyze circuit 3, \(I_D\) parallel with \(V_X=k_3I_D\).
This is again a current source in parallel with a voltage source, with the roles swapped from circuit 2. The voltage source freely supplies the current \(I_D\) that the current source demands, so this circuit is realizable for every value of \(k_3\).
Step 6: Analyze circuit 4, \(I_D\) parallel with \(I_X=3k_4I_D\).
Both arrows point the same way into the top node, so both currents try to enter the same node through the same two branches. With nothing else connected there, Kirchhoff's current law forces
\[
I_D+I_X=0
\]
so
\[
I_D+3k_4I_D=0 \implies k_4=-\frac{1}{3}.
\]
Step 7: Combine the results.
Circuit 1 fixes \(k_1=1\), circuit 4 fixes \(k_4=-\dfrac{1}{3}\), while circuits 2 and 3 place no restriction at all on \(k_2\) and \(k_3\).
Step 8: Final conclusion.
Therefore all four circuits are realizable only when
\[
\boxed{k_1=1,\ k_4=-\frac{1}{3},\ \text{for all values of }k_2\text{ and }k_3}
\]
Hence the correct option is (A).
