The calculation of the average power in ac circuits is presented with examples and their solutions. Problems with solutions are also included.
Consider the circuit below
Let the impedance \( Z \) be written in polar form
as \( Z = |Z| \; e^{j\theta} \)
Let \( v_i (t) = V_0 \; \cos(\omega t) \)
hence
\( i (t) = \dfrac{V_0}{|Z|} \; \cos (\omega t - \theta) \)
The instantenaous power \( P(t) \) delivered to the impedance \( Z \) is given by
\[ P(t) = i(t) \; v(t) = \dfrac{V_0^2}{|Z|} \; \cos ( \omega t - \theta) \; \cos(\omega t) \]
The average power is defined by
\[ P_a = \displaystyle \dfrac{1}{T} \int_0^T P(t) dt \]
Substitute \( P(t) \) by the expression found above and write the average power as
\( P_a = \displaystyle \dfrac{V_0^2}{T |Z|} \int_0^T \; \cos ( \omega t - \theta) \; \cos(\omega t) \; dt \)
Expand: \( \quad \cos ( \omega t - \theta) = \cos \omega t \; \cos \theta + \sin \omega t \; \sin \theta \) and substitute in \( P_a \)
\( P_a = \displaystyle \dfrac{V_0^2}{T |Z|} \int_0^T \; (\cos^2 \omega t \; \cos \theta + \sin \omega t \; \cos \omega t \; \sin \theta ) \; dt \)
Write the integral as a sum of two integrals: one integral on the left and a second integral on the right as follows:
\( P_a = \displaystyle \dfrac{V_0^2}{T |Z|} \int_0^T \; cos^2 \omega t \; \cos \theta \; dt + \dfrac{V_0^2}{T |Z|} \int_0^T \; \sin \omega t \; \cos \omega t \; \sin \theta \; dt \)
Using the trigonometric identity: \( \quad \sin(\omega t) \cos(\omega t) = \dfrac{1}{2} \sin(2 \omega t) \) to rewrite the integral on the right as
\( \displaystyle \dfrac{V_0^2}{T |Z|} \int_0^T \; \sin \omega t \; \cos \omega t \; \sin \theta \; dt = \dfrac{V_0^2}{2 T |Z|} \sin \theta) \int_0^T \sin (2 \omega t ) \; dt \)
\( \quad = - \displaystyle \dfrac{V_0^2}{2 T |Z|} \sin \theta \dfrac{1}{2 \omega } \left[\cos (2 \omega t ) \right]_0^T \)
\( \quad \quad = - \displaystyle \dfrac{V_0^2}{2 T |Z|} \sin \theta \dfrac{1}{2 \omega } \left[\cos 2 \omega T - \cos 0 \right] \)
Use the formula \( \quad \omega = \dfrac{2 \pi}{T} \) to simplify \( \cos 2 \omega T \)
\( \quad \quad \quad = - \displaystyle \dfrac{V_0^2}{2 T |Z|} \sin \theta \dfrac{1}{2 \omega } [\cos (4 \pi) - \cos 0] \)
\( \quad \quad \quad \quad = 0 \)
Use the trigonometric identity \( \quad \cos^2 \omega t = \dfrac{1}{2} (\cos(2 \omega t )+1) \) in the integral on the left and write
\( P_a = \displaystyle \dfrac{V_0^2}{2 T |Z|} \; \cos \theta \; \int_0^T \; (\cos(2 \omega t )+1) \; dt \)
Write the integral as a sum of two integrals
\( P_a = \displaystyle \dfrac{V_0^2}{2 T |Z|} \cos \theta \int_0^T \; \dfrac{1}{2} \cos(2 \omega t ) \; dt + \dfrac{V_0^2}{2 T |Z|} \; \cos \theta \; \int_0^T \; dt \)
In a similar way as above, it can be shown that \( \displaystyle \int_0^T \; \dfrac{1}{2} \cos(2 \omega t ) \; dt = 0 \)
Hence \( P_a \) is given by
\( \displaystyle P_a = \dfrac{V_0^2}{2 T |Z|} \; \cos \theta \; \int_0^T \; dt \)
\( \displaystyle \quad = \dfrac{V_0^2}{2 T |Z|} \; \cos \theta \left[t\right]_0^T \)
\[ \displaystyle \quad \quad P_a = \dfrac{V_0^2}{2 |Z|} \cos \theta \]
The term \( \cos \theta \) in the formula above is called the power factor.
Note that, in general, \( |Z| \) and \( \cos \theta \) depend on the frequency and therefore the average power depend on the frequency of the voltage (or current) source.
As shown above the calculations could be quite challenging and therefore a power calculator for series RLC circuit is included for more practice and investigations.
Example 1
In the series RLC circuit shown below, the source voltage is given by \( v_i = 5 \cos (\omega t) \), the capacitance of the capacitor \( C = 100 \; \mu F \), the inductance of the inductor \( L = 100 \; mH\) and the resistance of the resistor \( R = 1000 \; \Omega \) and frequency \( f = 2000 \; Hertz \).
a) Find the total impedance \( Z \) of the series RLC circuit and express it in polar form.
b) Find the average power delivered to the total impedance \( Z \).
Solution to Example 1
a)
For a series RLC circuit \( Z = R + j(\omega L - \dfrac{1}{\omega C} ) \)
\( \omega = 2 \pi f = 4000 \pi \) rad/s
Substitute \( R \), \( L \), \( C \) and \( \omega \) by their numerical values to obtain
\( Z = 1000 + j\left(4000 \pi \times 100 \times 10^{-3} - \dfrac{1}{4000 \pi \times 100 \times 10^{-6}} \right) \)
\( Z = 1000 + \left(400\pi -\dfrac{5}{2\pi} \right) j \)
The impedance \( Z \) is written in standard complex form \( Z = a + j b \)
In polar form the same impedance is written as \( Z = |Z| e^{j\theta} \)
where \( \theta = \arctan \dfrac{b}{a} \) and \( |Z| = \sqrt {a^2 + b^2} \)
Hence
\( \theta = \arctan \left(\dfrac{400\pi -\dfrac{5}{2\pi} }{1000} \right) \)
\( |Z| = \sqrt {1000^2 + \left(400\pi -\dfrac{5}{2\pi}\right)^2} \)
b)
\( \displaystyle P_a = \dfrac{V_0^2}{2 |Z|} \cos \theta \)
Substitute \( V_0 \), \( |Z| \) and \( \theta \) by their numerical values
\( \displaystyle P_a = \dfrac{5^2}{2 \sqrt {1000^2 + \left(400\pi -\dfrac{5}{2\pi}\right)^2} } \cos \left( \arctan \left(\dfrac{400\pi -\dfrac{5}{2\pi} }{1000} \right) \right) \)
\( \quad \approx 0.00485 \; \text{Watts} \)
Example 2
Show that the average power delivered to a series RLC circuit, such as the one in example 1, is maximum for a frequency \( f = \dfrac{1}{2 \pi \sqrt{LC}} \) and find a formula for this maximum power.
Solution to Example 2
a) For a series RLC circuit, the total impedance is given by: \( Z = R + j(\omega L - \dfrac{1}{\omega C}) \)
The angular frequency \( \omega \) is related to the frequency \( f \) by the formula: \( \omega = 2 \pi f = \dfrac{1}{\sqrt{LC}}\)
Substitute \( \omega \) by \( \dfrac{1}{\sqrt{LC}}\) in \( Z \) to obtain
\( Z = R + j (\dfrac{1}{\sqrt{LC}} L - \dfrac{1}{\dfrac{C}{\sqrt{LC}}}) \)
\( \quad = R + j ( \dfrac{1}{\sqrt{LC}} L - \dfrac{\sqrt{LC}}{C} ) \)
\( \quad = R + j ( \dfrac{\sqrt L}{\sqrt C} - \dfrac{\sqrt{L}}{\sqrt C} ) \)
which simplifies to
\( Z = R \)
For the frequency \( f = \dfrac{1}{2 \pi \sqrt{LC}} \), the impedance \( Z \) is real and therefore
\( |Z| = R \)
and the argument \( \theta \) of \( Z \) is equal to zero. Hence \( \cos \theta = \cos 0 = 1 \) has a maximum value.
For the frequency \( f = \dfrac{1}{2 \pi \sqrt{LC}} \), the power factor \( \cos \theta = \cos 0 = 1 \) is maximum and \( |Z| \) is minimum which gives an average power with a maximum value given by
\( P_a max = \dfrac{V_0^2}{2 R} \)
Example 3
In a series RLC circuit, such as the one in example 1 above, the source voltage is given by \( v_i = 2 \cos ( \omega t) \), the capacitance of the capacitor \( C = 470 \mu \)F, the inductance of the inductor \( L = 50 \)mH and the resistance of the resistor is \( R = 100 \; \Omega \).
a) Express the average power \( P_a \) given by its formula above and make a graph of \( P_a \) as a function of the angular frequency \( \omega \) and find the location of the maximum of \( P_a \).
b) Check that the power is maximum at the angular \( \omega_r = \dfrac{1}{\sqrt{LC}} \) (or the frequency \( f_r = \dfrac{1}{2 \pi \sqrt{LC}} \)) and is given by \( P_a max = \dfrac{V_0^2}{2 R} \) as explained in example 2 above.
Solution to Example 3
For a series RLC circuit \( Z = R + j(\omega L - \dfrac{1}{\omega C}) \)
Modulus: \( |Z| = \sqrt { R^2 + \left( \omega L - \dfrac{1}{\omega C} \right)^2} \)
Argument: \( \theta = \arctan \left( \dfrac{\omega L - \dfrac{1}{\omega C}} {R} \right) \)
The average power \( P_a \) is given by
\[ P_a = \dfrac{V_0^2}{2 |Z|} \cos \theta \]
Substitute \( V_0 \) and \( |Z| \) by their expressions
\( P_a (\omega ) = \dfrac{V_0^2}{2 \sqrt { R^2 + \left( \omega L - \dfrac{1}{\omega C} \right)^2}} \cos \left(\arctan \left( \dfrac{\omega L - \dfrac{1}{\omega C}} {R} \right) \right) \)
Substitute \( R \), \( L \) and \( C \) by their values to obtain \( P_a \) as a function of \( \omega \)
\( P_a (\omega ) = \dfrac{2}{ \sqrt { 100^2 + \left( 50 \times 10^{-3} \; \omega - \dfrac{1}{470 \times 10^{-6}\; \omega } \right)^2}} \cos \left(\arctan \left( \dfrac{50 \times 10^{-3} \; \omega - \dfrac{1}{ 470 \times 10^{-6} \; \omega }} {100} \right) \right) \)
The graph of \( P_a (\omega ) \) against \( \omega \) is shown below.
The free geogebra graphing calculator was used to graph and locate the maximum as shown in the graph.
b)
In example 2 above, it was explained the average power is maximum at
\( \omega_r = \dfrac{1}{\sqrt{LC}} = \dfrac{1}{ \sqrt{50 \times 10^{-3} \times 470 \times 10^{-6} }} \approx 206.28\) rad/s
The maximum power is given by: \( P_a max = \dfrac{V_0^2}{2 R} = \dfrac{2^2}{2 \times 100} = 0.02\) Watts
Both the calculated values of \( \omega_r \) and \( P_a max \) calculated are equal to values found graphically above.