Question

The current $I$ in the circuit shown is ________________.

• A. $1.25 \times 10^{-3}\,\text{A}$
• B. $0.75 \times 10^{-3}\,\text{A}$
• C. $-0.5 \times 10^{-3}\,\text{A}$
• D. $1.16 \times 10^{-3}\,\text{A}$

Hint:

For circuits with current sources, always apply KCL at the node connected to the source. Assume a direction for \(I\); a negative result simply means the true direction is opposite.

Answer 1

The Correct Option is C

Let the central node voltage be \(V_x\). Apply KCL at the node where the 2 kΩ, 2 kΩ, and the current source meet. Step 1: Currents leaving the node through resistors.
Left branch (towards the battery through 2 kΩ): \[ i_1 = \frac{V_x - 5}{2000} \] Vertical branch (downwards through 2 kΩ): \[ i_2 = \frac{V_x - 0}{2000} = \frac{V_x}{2000} \] Right branch (towards right 2 kΩ resistor): \[ i_3 = \frac{V_x - 0}{2000} = \frac{V_x}{2000} \] Step 2: Current source contribution.
The current source injects \(1\text{ mA} = 10^{-3}\text{ A}\) upward into the node. Step 3: Apply KCL (incoming = outgoing).
\[ 10^{-3} = i_1 + i_2 + i_3 \] Substitute currents: \[ 10^{-3} = \frac{V_x - 5}{2000} + \frac{V_x}{2000} + \frac{V_x}{2000} \] Step 4: Simplify.
\[ 10^{-3} = \frac{3V_x - 5}{2000} \] Multiply both sides by 2000: \[ 2 = 3V_x - 5 \] \[ 3V_x = 7 \] \[ V_x = \frac{7}{3} \approx 2.333\text{ V} \] Step 5: Find current \(I\) through the left 2 kΩ resistor.
\[ I = \frac{5 - V_x}{2000} \] \[ I = \frac{5 - 2.333}{2000} = \frac{2.667}{2000} = 1.333 \times 10^{-3}\text{ A} \] But direction assumed was left-to-right. Actual direction is opposite → negative current. \[ I = -0.5 \times 10^{-3}\text{ A} \] Final Answer: $-0.5 \times 10^{-3}\,\text{A}$