Using the superposition principle, find the voltage across the 1 kΩ resistor in the figure. Assume the sources to be ideal.

Steps to Solve

1. Analyze Each Independent Source To apply the superposition principle, analyze the circuit for each independent source separately. There are two independent sources: the 10 mA current source and the 15 V voltage source.

2. Contribution from the 10 mA Current Source When considering the 10 mA current source alone:

• Remove the 15 V source (replace with a short circuit).

• The voltage across the 1 kΩ resistor can be calculated using Ohm's Law and the total current flowing through the resistor.

  The circuit can be simplified to the 10 mA source feeding into the 4 kΩ and 3 kΩ resistors in parallel, and then into the 1 kΩ resistor.

3. Calculate Equivalent Resistance The equivalent resistance of the parallel resistors is: $$ R_{\text{eq}} = \frac{1}{\frac{1}{4000} + \frac{1}{3000}} = 1200 , \Omega $$

4. Calculate the Total Current through the 1 kΩ Resistor Using the equivalent resistance: $$ I = \frac{10 , \text{mA}}{1 + \frac{R_{\text{eq}}}{1000}} = \frac{10}{1 + \frac{1200}{1000}} , \text{mA} = 4.167 , \text{mA} $$

5. Calculate Voltage across the 1 kΩ Resistor Using Ohm’s Law ($V = IR$): $$ V_{1k\Omega} = 4.167 , \text{mA} \times 1000 , \Omega = 4.167 , V \text{ (from the current source)} $$

6. Contribution from the 15 V Voltage Source Next, analyze the contribution of the 15 V source:

• Remove the current source (replace with an open circuit).
• The voltage across the 1 kΩ resistor can be evaluated using the voltage divider rule.

7. Calculate the Voltage across the Resistors Using voltage division: $$ V_{1k\Omega} = 15 , V \times \frac{1k\Omega}{1k\Omega + 4k\Omega + 3k\Omega} = 15 , V \times \frac{1}{8} = 1.875 , V \text{ (from the voltage source)} $$

8. Combine Contributions Finally, sum the voltages from both sources to find the total voltage across the 1 kΩ resistor: $$ V_{\text{total}} = V_{\text{current source}} + V_{\text{voltage source}} = 4.167 , V + 1.875 , V = 6.042 , V $$