Name: Power in AC Circuit
Uploaded: 2019-07-19
Duration: 4 min 36 s
Description: The video defines the expression for the instantaneous power in an ac circuit. It also derives the expression for the average power consumed by an AC circuit.

Question

Are the readings of the voltmeter $V 3$ equal to the sum of the readings of the voltmeters $V 1$ and $V 2$ for the cases depicted in Figures ?

Accepted Answer

For the case depicted in Figure $(a)$ accompanying the problem, the voltage between points $1$ and $2$ is

$U_{1} = I_{0} R \sin ω t = U_{01} \sin ω t$ ,

while that between points $2$ and $3$ is

$U_{2} = \frac{I_{0}}{C ω} \sin (ω t - \frac{π}{2}) = - \frac{I_{0}}{C ω} \cos ω t = - U_{02} \cos ω t$

(see the vector diagram in Figure $(a)$ accompanying the answer). The voltage between points $1$ and $3$ is the sum of $U_{1}$ and $U_{2}$ :

$U_{3} = U_{1} + U_{2} = U_{03} \sin (ω t + φ)$ .

Its amplitude value is

$U_{03} = \sqrt{U_{01}^{2} + U_{02}^{2}}$ ,

while the phase shift with respect to the applied voltage is given by the following formula:

$\tan φ = - \frac{U_{02}}{U_{01}}$ .

Since the voltmeters measure the effective value $U_{e f f} =$ $\frac{U_{0}}{\sqrt{2}}$ , we have $U_{3 e f f} = \sqrt{U_{1 e f f}^{2} + U_{2 e f f}^{2}} < U_{1 e f f} + U_{2 e f f}$ .

For the circuit depicted in Figure $(b)$ accompanying the problem, just like in the previous case, the voltages across the resistance and the inductance differ in phase by $\frac{π}{2}$ , the only difference being that here the current flowing through the inductance lags behind the voltage, while in the previous case the current flowing through the capacitance leads the voltage (and, hence, the phase shift between the voltages across the resistance and across the capacitance is $\frac{- π}{2}$ ). The respective voltages are

$U_{1} = I_{0} R \sin ω t = U_{01} \sin ω t$ ,

$U_{2} = I_{0} L ω \sin (ω t + \frac{π}{2}) = U_{02} \cos ω t$

and $U_{3} = U_{3} \sin (ω t + φ)$

(see the vector diagram depicted in Figure $(b)$ accompanying the answer). The amplitude value of the voltage is

$U_{03} = \sqrt{U_{01}^{2} + U_{02}^{2}}$ .

The effective voltages measured by the voltmeters are related thus:

$U_{3 eff} = \sqrt{{\bar{U}}_{1 eff}^{2} + U_{2 eff}^{2}} < U_{1 eff} + U_{2 eff}$ .

The tangent of the phase of the voltages is

$\tan φ = \frac{U_{02}}{U_{01}}$ .

L.A.Sena Solutions for Chapter: Alternating Current, Exercise 1: Exercise

L.A.Sena Physics Solutions for Exercise - L.A.Sena Solutions for Chapter: Alternating Current, Exercise 1: Exercise

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