Since ac circuits are linear, the superposition theorem applies to ac circuits the same way it applies to superposition theorem dc circuits.

## Superposition Theorem AC Circuit

The theorem becomes important if the circuit has sources operating at

*different*frequencies. In this case, since the impedances depend on frequency, we must have a different frequency domain circuit for each frequency.
The total response must be obtained by adding the individual responses in the

*time*domain. It is incorrect to try to add the responses in the phasor or frequency domain. It is because the exponential factor e^{jωt}is implicit in sinusoidal analysis, and that factor would change for every angular frequency ω. It would therefore not make sense to add responses at different frequencies in the phasor domain. Hence, when a circuit has sources operating at different frequencies, one must add the responses due to the individual frequencies in the time domain.## Superposition Theorem AC Circuit Examples

For better understanding, let us review the examples below :

**1. Use the superposition theorem to find Io in the circuit in Figure.(1).**

Figure 1 |

*Solution :*
Let

(1.1) |

where

**I**'_{o}and**I**"_{o}are due to the voltage and current sources, respectively. To find**I**'_{o}, consider the circuit in Figure.(2a).Figure 2 |

If we let

**Z**be the parallel combination of -*j*2 and 8 +*j*10, then
and current

**I**'_{o}is
or

(1.2) |

To get

**I**"_{o}, consider the circuit in Figure.(2b). For mesh 1,(1.3) |

For mesh 2,

(1.4) |

For mesh 3,

(1.5) |

From Equations.(1.4) and (1.5),

Expressing

**I**_{1}in terms of**I**_{2}gives(1.6) |

Substituting Equations.(1.5) and (1.6) into Equation.(1.3), we get

or

Current

**I**"_{o}is obtained as(1.7) |

From Equations.(1.2) and (1.7), we write

**2. Find v**

_{o}in the circuit of Figure.(3) using the superposition theorem.Figure 3 |

*Solution :*
Since the circuit operates at three different frequencies (ω = 0 for the dc voltage source), one way to obtain a solution is to use superposition, which breaks the problem into single-frequency problems. So we let

(2.1) |

where v

_{1}is due to the 5 V dc voltage source, v_{2}is due to the 10 cos 2t V voltage source, and v_{3}is due to the 2 sin 5t A current source.
To find v

_{1}, we set to zero all sources except the 5 V dc source. We recall that at steady state, a capacitor is an open circuit to dc while an inductor is short circuit to dc. There is an alternative way of looking at this. Since ω = 0, jωL = 0, 1/jωC = ∞. Either way, the equivalent circuit is as shown in Figure.(4a). By voltage division,(2.2) |

To find v

_{2}, we set to zero both the 5 V source and the 2 sin 5t current source and transform the circuit to the frequency domain.
The equivalent circuit is now as shown in Figure.(4b).Let

By voltage division,

In the time domain,

(2.3) |

To obtain v

_{3}, we set the voltage sources to zero and transform what is left to the frequency domain.
The equivalent circuit is in Figure.(4c). Let

By current division,

In the time domain,

(2.4) |

Substituting Equations.(2.2) to (2.4) into Equation.(2.1), we have

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