Question

Find the value of current \(I\) in the circuit shown in the figure below.

• A. \(3\,A\)
• B. \(13\,A\)
• C. \(6\,A\)
• D. \(-13\,A\)

Hint:

At every junction, \[ \sum I_{\text{entering}} = \sum I_{\text{leaving}}. \] Solve junction currents one by one using KCL.

Answer 1

The Correct Option is B

Concept: Apply Kirchhoff's Current Law (KCL): \[ \sum I_{\text{in}}=\sum I_{\text{out}}. \]

Step 1: Consider the upper-left junction. Incoming current: \[ 15\,A \] Outgoing currents: \[ 8\,A \] and current in the upper branch \(I_t\). Therefore, \[ 15=8+I_t \] \[ I_t=7\,A. \]

Step 2: Consider the lower-left junction. Incoming current: \[ 8\,A. \] Outgoing currents: \[ 5\,A \] and lower branch current \(I_b\). Thus, \[ 8=5+I_b \] \[ I_b=3\,A. \]

Step 3: Consider the upper-right junction. Incoming current: \[ 7\,A+3\,A=10\,A \] must leave through the right vertical branch together with the external branch. Since \(3\,A\) enters the junction from outside, \[ I_{\text{right vertical}} = 7+3 = 10\,A. \]

Step 4: Apply KCL at the lower-right junction. Incoming currents: \[ 10\,A+3\,A=13\,A. \] Therefore the outgoing current is \[ I=13\,A. \] \[\begin{aligned} \boxed{I=13\,A} \end{aligned}\] Hence, option \(\mathbf{(B)}\) is correct.