Question

Find the average power consumed in a series A.C. circuit.
A) ${E_{rms}}{I_{rms}}\cos \phi $
B) ${\left( {{I_{rms}}} \right)^2}R$
C) $\dfrac{{E_0^2}}{{2{{\left( {\left| z \right|} \right)}^2}}}$
D) $\dfrac{{I_0^2\left| z \right|\cos \phi }}{2}$

Answer 1

Hint:The power consumed or dissipated in a circuit, with the current $I$ in the circuit and the voltage $E$across the circuit will be in the form of $P = IE$ . The average power refers to the time average of the instantaneous power over one cycle.

Formulas used:
-The average power consumed in a series A.C. circuit is given by, ${P_{av}} = \dfrac{1}{T}\int\limits_0^T {p\left( t \right)dt} $ where $p\left( t \right)$ is the instantaneous power.
-Instantaneous power in the circuit is given by, $p\left( t \right) = i\left( t \right)E\left( t \right)$ where $i\left( t \right)$ and $E\left( t \right)$ represent the instantaneous current and voltage in the circuit.

Complete step by step answer.
Step 1: Describe the parameters of the a.c. circuit under consideration.
In an a.c. circuit, the current and the voltage depend on time.
We represent instantaneous current in the circuit as $i\left( t \right) = {I_0}\sin \left( {\omega t - \phi } \right)$ and the instantaneous voltage across the circuit as $E\left( t \right) = {E_0}\sin \omega t$
Here, $\omega = \dfrac{{2\pi }}{T}$ is the angular frequency and $\phi $ is the phase angle which is the phase difference between the current and voltage in the circuit. Also $T$ is the period of oscillation.
And, ${I_0}$ and ${E_0}$ refers to the amplitude of the current and voltage respectively.
Step 2: Obtain an expression for the instantaneous power in the circuit.
The instantaneous power in the circuit is given by, $p\left( t \right) = i\left( t \right)E\left( t \right)$ .
Substituting the $i\left( t \right) = {I_0}\sin \left( {\omega t - \phi } \right)$ and $E\left( t \right) = {E_0}\sin \omega t$ in the above equation we have, $p\left( t \right) = {I_0}{E_0}\sin \omega t\sin \left( {\omega t - \phi } \right)$
Thus the instantaneous power in the circuit is $p\left( t \right) = {I_0}{E_0}\sin \omega t\sin \left( {\omega t - \phi } \right)$ -------- (1).
Step 3: Using equation (1) find the average power consumed in the a.c. circuit.
The average power consumed in an a.c. circuit is the time average of the instantaneous power.
i.e., ${P_{av}} = \dfrac{1}{T}\int\limits_0^T {p\left( t \right)dt} $ ---------- (2)
Substituting equation (1) in the (2) we get, ${P_{av}} = \dfrac{1}{T}\int\limits_0^T {{I_0}{E_0}\sin \omega t\sin \left( {\omega t - \phi } \right)dt} $
Or, we get, ${P_{av}} = \dfrac{{{I_0}{E_0}}}{T}\int\limits_0^T {\sin \omega t\sin \left( {\omega t - \phi } \right)dt} $
We use the trigonometric identity, $\sin \left( {a - b} \right) = \sin a\cos b - \sin b\cos a$ , to split the above integral.
Now we have, ${P_{av}} = \dfrac{{{I_0}{E_0}\cos \phi }}{T}\int\limits_0^T {{{\sin }^2}\omega tdt} - \dfrac{{{I_0}{E_0}\sin \phi }}{T}\int\limits_0^T {\sin \omega t\cos \omega t} dt$ ------- (3)
Here, the terms $\sin \phi $ and $\cos \phi $ do not depend on time.
The individual integrals are obtained as $\dfrac{1}{T}\int\limits_0^T {{{\sin }^2}\omega tdt} = \dfrac{1}{2}$ and $\int\limits_0^T {\sin \omega t\cos \omega t} dt = 0$
We now substitute the values of the individual integrals in equation (3).
Then the average power becomes, ${P_{av}} = \dfrac{{{I_0}{E_0}\cos \phi }}{2}$
For a resistive circuit, the phase angle is zero i.e., $\phi = 0$ and thus $\cos \phi = 1$ .
Thus we have, ${P_{av}} = \dfrac{{{I_0}{E_0}}}{2}$ --------- (4).
Now we express the above relation in terms of the rms values of the current and voltage.
The rms value of the current is defined as ${I_{rms}} = \sqrt {i_{av}^2} = \dfrac{1}{{\sqrt 2 }}{I_0}$ ------- (A)
Similarly, the rms value of the voltage is defined as ${E_{rms}} = \sqrt {E_{av}^2} = \dfrac{1}{{\sqrt 2 }}{E_0}$ ------- (B)
From equation (A) we get, ${I_0} = \sqrt 2 {I_{rms}}$ and from equation (B) we have ${E_0} = \sqrt 2 {E_{rms}}$
Substituting the above values for ${I_0}$ and ${E_0}$ in equation (4) we get, ${P_{av}} = {I_{rms}}{E_{rms}}$
Since ${E_{rms}} = {I_{rms}}R$ we finally have ${P_{av}} = I_{rms}^2R$

Therefore, the average power consumed in an a.c. circuit is ${P_{av}} = I_{rms}^2R$ .

Note: The magnitude, as well as the sign of the instantaneous power, varies in one cycle. This is why we always determine the average power consumed in a circuit. The phase angle is the angle by which the current lags behind or leads before the voltage. So, if there exists a phase difference between the current and voltage then they will not attain their maximum value at the same instant.