Superposition Example

The circuit shown below provides an illustration of the principle of superposition. The values of the voltage source voltage and current source current and of both resistances can be adjusted using the scrollbars. Also, both of the sources can be removed form the circuit and put back into the circuit by left-clicking the source with the mouse.

The voltage source voltage and current source current are the inputs to this circuit. The voltage and current measured by the meters are the outputs or responses of the circuit.The principle of superposition says that the response of a linear circuit to several inputs working together is equal to the sum of the response of that circuit to the inputs working separately.

In this example, let

1. I and V denote the current and voltage measured by the meters when both the voltage source and the current source are in the circuit. That is, I and V, are both responses to Vs and Is working together.

2. I1 and V1 denote the current and voltage measured by the meters when the voltage source, but not the current source, is in the circuit. That is, I1 and V1, are both responses to Vs working alone.

3. I2 and V2 denote the current and voltage measured by the meters when the current source, but not the voltage source, is in the circuit. That is, I2 and V2, are both responses to Is working alone.

The principle of superposition tells us that

For example, use the scrollbars to set Is= 4 A, Vs = 20 V, R1=10 Ohms and R2=30 ohms.

1. The meters indicate that I=-2.5 A and V=45 V.

2. Left-click on the current source to remove it from the circuit. (Notice that the current source is replaced by an open circuit because an open is equivalent to a zero current source.) Now the meters measure I1 and V1 because the voltage source, but not the current source, is in the circuit.The meters indicate that I1=0.5 A and V1=15 V.

3. Left-click on the current source to restore it to the circuit. Once again the meters measure I and V because both the voltage source and the current source are in the circuit. As expected, the meters gain indicate that I=-2.5 A and V=45 V.

4. Left-click on the voltage source to remove it from the circuit. (Notice that the voltage source is replaced by a short circuit because a short is equivalent to a zero voltage source.) Now the meters measure I2 and V2 because the current source, but not the voltage source, is in the circuit.The meters indicate that I2=-3 A and V2=30 V.

5. The principle of superposition predicts that . Indeed, this is the case:

Challenges:

1. Use the scrollbars to set Is= 1 A, Vs = 40 V, R1=10 Ohms and R2=40 Ohms.

1. Observe the values of I and V.

2. Left-click on the current source to remove it from the circuit. Notice that R1 and R2 are series resistors and that V1 can be calculated using voltage division. (See section 3.4 of Introduction to Electric Circuits, 5e by R.C. Dorf and J.A. Svoboda.) Observe the values of I1 and V1.

3. Left-click on the current source to restore it to the circuit then left-click on the voltage source to remove it from the circuit. Notice that R1 and R2 are parallel resistors and that I2 can be calculated using current division. (See section 3.5 of Introduction to Electric Circuits by RC Dorf and JA Svoboda.) Observe the values of I2 and V2.

4. Verify that .

2. In Example 5.4-1 of Introduction to Electric Circuits, 5e by R.C. Dorf and J.A. Svoboda, Is= 2 A, Vs = 6 V, R1=3 Ohms and R2=6 Ohms. Example 5.4-1 shows that the current in R2 is 4/3 A. Equivalently, the voltage across R2 is 6 * 4/3 = 8 V. Use the scrollbars to verify this answer.

3. Use the scrollbars to set Is= 3 A, R1=10 Ohms and R2=5 Ohms. Predict the value of Vs required to make V = 25 V. Use the scrollbar to check your prediction.

4. Use the scrollbars to set Is= 0.6 A, Vs = 28 V. Predict the value of R1 = R2 required to make V = 20 V. Use the scrollbars to check your prediction.

5. Use the scrollbars to set R1 = R2 = 40 Ohms. Predict the value of Is and Vsand required to make V = 20 V and I = 0.1 A. Use the scrollbars to check your prediction.

(Hint: Use the principle of superposition to show that .)