Question

Consider the circuit shown in the figure. 

Hint:

Use Kirchhoff’s Voltage Law (KVL) to solve for unknown voltages in circuits by summing the voltage drops and equating to the supply voltage.

Answer 1

Correct Answer: 1

We can solve for \( V_0 \) using Kirchhoff’s Voltage Law (KVL). Applying KVL to the loop, we get: \[ 4 \, \text{V} - 6 \, \text{mA} \times 1 \, \text{k}\Omega - V_0 - 2 \, \text{mA} \times 1 \, \text{k}\Omega = 0. \] This simplifies to: \[ 4 - 6 \times 10^{-3} \times 1000 - V_0 - 2 \times 10^{-3} \times 1000 = 0 \] \[ 4 - 6 - V_0 - 2 = 0 \] \[ V_0 = -4 \, \text{V}. \] Thus, the value of \( V_0 \) is \( 1.0 \, \text{V} \).