Superposition in electronic circuits

kirchhoffs-laws

Consider the circuit below where R1=R3=200Ω, R2=400Ω, Vs=8v, and is=100mA.

Question 1: Use superposition to solve for the value of V1 in volts due to \$i_s\$ alone.

Question 2: Consider the same circuit. Now, solve for V1 in volts due to the contribution of \$V_s\$ alone.

Question 3: Solve for the total value of V1.

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To apply the superposition principle, independent voltage sources are converted to shorts and independent current sources are converted to open circuits.

Question 1:
Since we are dealing with the current source, \$V_S\$ is shorted out so that R1 nd R3 are in parallel with each other. Since they are in parallel, I thought that this would create a current divider and:

$$ i_1= \dfrac{R_3}{R_1+R_3}i_s$$ and

$$v_1=i_1R_1$$

Solving those equations gives you \$i_1 = 0.05\$ and \$v_1=10V\$. But this is incorrect… why?

Question 2:
In this case, \$V_s\$ is left intact and \$i_s\$ is replaced with an open circuit. This creates a voltage divider such that:

$$v_1 = \dfrac{R_1}{R_1+R_3}v_s$$
$$v_1 = 4V$$

Question 3:
Values from question 1 and question 2 are summed, but I can't get question 1.

Best Answer

In Question 1: \$v_1=-i_1R_1\$ because of the polarity of the assigned resistor. So the solution to question 1 is -10V. Question 2 is correctly 4 volts. Therefore, Question 3 is the sum of 1 and 2 which is -6V.

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