Question 1
In the circuit shown in the figure , the power supplied by the voltage source is:

Solution
- Method: Apply Kirchhoff's Voltage Law (KVL) to the outer loop to find the current (\(I_1\)) supplied by the \(10\,V\) source.
- Branch Currents: Let \(I_1\) be the current leaving the \(10\,V\) source. The two current sources fix the additional current carried by each arm of the outer loop, so the two \(2\,\Omega\) arms carry \(I_1 + 3\) and \(I_1 + 2\).
- KVL on the Outer Loop: The drops produced by the two current sources already account for the whole \(10\,V\), so the source itself carries no current.Equation\[2(I_1 + 3) + 2(I_1 + 2) = 10 \Rightarrow 4I_1 + 10 = 10 \Rightarrow I_1 = 0\,A\]
- Power Calculation: Power supplied by the voltage source is \(P = V \times I_1\). Equation\[P = 10\,V \times 0\,A = 0\,W\]
Final Answer
Correct answer: (1) \(0\,W\).