Alternating Current and Voltages change their magnitude and direction with time. This changes the way calculations for power and other quantities are done in circuits. Furthermore, with the introduction of capacitors and inductances, many other effects come into play which alters the power calculation in these circuits from the usual ways of calculating power with DC sources. It becomes essential to know these workings and concepts so that one can apply them in situations in real life where circuits are complex and power calculation is required. Let’s look at these concepts in detail.

### Power in AC Circuits

Even though the average current through the cycle is zero, but that does not mean that average power dissipation through the cycle is also zero. Dissipation of electrical energy is there. It’s known that Joule’s heating is given by i^{2}R and depends on i^{2}. This term is always positive irrespective of the sign of “i”. Thus average dissipation cannot become zero. Let us consider a circuit that is connected to an alternating voltage source. Its voltage expression is given below,

v = v_{m}sinÏ‰t

In case the circuit is given an RLC circuit, the voltage source drives a current in the system which is given by,

i = i_{m}sin(Ï‰t + Ï†)

Here,

i_{m} = and

In this case, the instantaneous power supplied by the source becomes,

p = vi

â‡’ p = (v_{m}sinÏ‰t)(i_{m}sin(Ï‰t + Ï†))

â‡’ p =

This is the instantaneous power, this means this is the value of the power being dissipated in the circuit at any given point of time. To calculate the average power dissipated, we need take average of the both of the terms given in the equation above.

Notice that here, only the second term is time-dependent and it’s average is zero. This leaves the first term, this term is gives us the non-zero average power.

p =

â‡’ p =

This can also be written using the rms values of the voltage and current.

P = VI cos(Ï†)

This can also be rewritten as,

P = I^{2}Zcos(Ï†)

The above derivation proves that the average power dissipated in the circuit depends not only on current and voltage. It also depends on the cosine of their phase angle. The quantity cos(Ï†) in this case is called the** power factor. **Based on the derivation given above, three cases can be considered:

- Resistive Circuit
- Purely Inductive or Capacitive Circuit
- LCR series circuit
- The power

**Resistive Circuit **

A resistive circuit only contains resistances, in such cases, the calculation for power becomes fairly simple because there is no phase difference between the current and the voltage vectors. So, in that case, Ï† = 0. Plugging the value of this angle in the equation for power derived above,

P = VI cos(Ï†)

â‡’ P = VI cos(0)

â‡’ P = VI

In this case, the power dissipation is maximum.

**Purely Inductive or Capacitive Circuit **

In this case, the circuit contains only a capacitor or inductor. The phase difference between voltage and current, in this case, will be 90Â°. So, the angle Ï† = 90Â°. Plugging the value of this angle in the equation for power derived above,

P = VI cos(Ï†)

â‡’ P = VI cos(90)

â‡’ P = 0

In this case, the power dissipation is minimum.** **

**LCR series circuit**

In this case, the circuit contains all the elements resistor, capacitor, and inductance. the angle Ï†, in this case, is given by

Ï† =

This is a generalized version of the previous two cases, the power factor can be zero or non-zero.

The power dissipated at resonance in the LCR circuit: At resonance X

_{C }– X_{L}= 0 and Ï† = 0. In this case, cos(Ï†) = 1. Thus, the power dissipated in this case is maximum.

### Sample Problems

**Question 1: A sinusoidally varying current is applied to a purely resistive circuit. The resistance is 20 ohms and the RMS value of the voltage is 10V. Find the power dissipated in the circuit. **

**Answer: **

The Average power is given by,

P = VIcos(Ï†)

Since the circuit is purely resistive circuit.

Ï† = 0

cos(Ï†) = 1

P = V

^{2}/Rplugging the values in the given formula.

P = (10)

^{2}/20â‡’ P = 100 /20

â‡’ P = 5W

**Question 2: A sinusoidally varying current is applied to a purely resistive circuit. The resistance is 10 ohms and the RMS value of the voltage is 30V. Find the power dissipated in the circuit. **

**Answer: **

The Average power is given by,

P = VIcos(Ï†)

Since the circuit is purely resistive circuit.

Ï† = 0

cos(Ï†) = 1

P = V

^{2}/RPlugging the values in the given formula.

P = (30)

^{2}/10â‡’ P = 900 /20

â‡’ P = 45W

**Question 3: A sinusoidally varying current is applied to an LC circuit. The impedance of the inductor and capacitance is given as 8 and 2 ohms. Find the power dissipated in the circuit if the voltage source has an RMS voltage of 45V. **

** Answer: **

The Average power is given by,

P = VIcos(Ï†)

Since the circuit is purely capacitive and inductive circuit. The phase angle will be 90Â°.

cos(Ï†) = 0

Plugging the values in the equation,

P = VIcos(Ï†)

P = 0.

**Question 4: A sinusoidally varying current is applied to an LCR circuit. The impedance of the inductor and capacitance is given as 2 and 8 ohms** **while the resistance is 6 ohms. Find the power dissipated in the circuit if the voltage source has an RMS voltage of 45V and the RMS value of the current flowing through the circuit is given by 20A. **

** Answer: **

The Average power is given by,

P = VIcos(Ï†)

Since the circuit is LCR circuit. The phase angle will be calculated by the given formula,

Ï† =

â‡’ Ï† =

â‡’ Ï† = 45Â°

Plugging the values in the equation,

P = VIcos(Ï†)

P = (45)(20)cos(45)

â‡’ P = (900)cos(45)

â‡’ P = 450âˆš2 W.

**Question 5: A sinusoidally varying current is applied to an LCR circuit. The impedance of the inductor and capacitance is given as 2 and 8 ohms** **while the resistance is 8 ohms. Find the power dissipated in the circuit if the voltage source has an RMS voltage of 90V and the RMS value of the current flowing through the circuit is given by 10A. **

** Answer: **

The Average power is given by,

P = VIcos(Ï†)

Since the circuit is LCR circuit. The phase angle will be calculated by the given formula,

Ï† =

â‡’ Ï† =

â‡’ Ï† = 37Â°

plugging the values in the equation,

P = VIcos(Ï†)

P = (90)(10)cos(37)

â‡’ P = (900)cos(37)

â‡’ P = (900 Ã— 4)/5

â‡’ P = 720W.