Question 17.8: Determine the average power supplied to the circuit in Fig. ......

The input impedance of the network is

Z = 10 \parallel \frac{1}{j2ω} = \frac{10(1/j2ω)}{10  +  1/j2ω} = \frac{10}{1  +  j20 ω}

Hence,

V = IZ = \frac{10 I}{\sqrt{1  +  400ω²} \underline{/ \tan^{-1} 20 ω}}

For the dc component , ω = 0 ,

I =  2 A      ⇒           V = 10(2)  =  20 V

This is expected, because the capacitor is an open circuit to dc and the entire 2-A current flows through the resistor. For ω = 1  rad/s ,

I = 10 \underline{/ 10°}    ⇒             V = \frac{10(10 \underline{/ 10°})}{\sqrt{1  +  400} \underline{/ \tan^{-1} 20 }}

= 5\underline{/ -77.14°}

For ω  =  3 rad/s ,

I = 6 \underline{/ 35°}    ⇒             V = \frac{10(6 \underline{/ 35°})}{\sqrt{1  +  3600} \underline{/ \tan^{-1} 60 }}

= 1\underline{/ -54.04°}

Thus, in the time domain,

v(t) =  20   + 5  \cos(t  –  77.14°)   + 1 \cos(3t  –  54.04°) V

We obtain the average power supplied to the circuit by applying Eq. (17.46), as

P =  V_{dc} I_{dc}+ \frac{1}{2}  \sum\limits^{∞}_{n = 1}{V_{n}I_{n}  \cos(θ_{n}  –  Φ_{n})}\quad \quad (17.46)

To get the proper signs of  θ_{n}   and  Φ_{n}   we have to compare v and i in this example with Eqs. (17.42) and (17.43). Thus,

v(t) =  V_{dc}  +  \sum\limits_{n = 1}^{∞}{V_{n}  \cos(nω_{0}t   –  θ_{n})}                                     (17.42)

i(t) =  I_{dc}   +  \sum\limits^{∞}_{m = 1}{I_{m}  \cos(mω_{0}t  – Φ_{m})}                                        (17.43)

P= 20(2)  +  \frac{1}{2} (5)(10) \cos [77.14°  –  (-10°)]

+ \frac{1}{2} (1)(6) \cos [54.04°  –  (-35°)]

= 40 + 1.247  +  0.05  =  41.5  W

Alternatively, we can find the average power absorbed by the resistor as

P = \frac{V^{2}_{dc}}{R}  +  \frac{1}{2}  \sum\limits^{∞}_{n = 1}{\frac{\left|V_{n}\right|^{2} }{R}} = \frac{20²}{10}  +  \frac{1}{2}   . \frac{5²}{10}  + \frac{1}{2} . \frac{1²}{10}

=  40   +  1.25  +   0.05  =  41.5  W

which is the same as the power supplied, since the capacitor absorbs no average power.