Suppose the circuit in Question 7.18 has a resistance of 15 Ω Obtain the average power transferred to each element of the circuit, and the total power absorbed.

Solution

Here,
Reistance, $R = 15 Ω$
$∴$ Impedance, $Z = \sqrt{R^{2} + {(ωL - \frac{1}{ωC})}^{2}}$

$Z = \sqrt{15^{2} + {(2 π \times 50 \times 80 \times 10^{- 3} - \frac{1}{2 π \times 50 \times 60 \times 10^{- 6}})}^{2}}$
$= \sqrt{225 + 779.5} Ω = 31.7 Ω$

$∴$ $I_{rms} = \frac{V_{rms}}{Z} = \frac{230}{31.7} = 7.255 A$

Average power transferred to inductor, $L = E_{v} I_{v} \cos \frac{π}{2} = 0$
Average power transferred to capacitor= $E_{v} I_{v} \cos (\frac{- π}{2}) = 0$
Average power transferred to R = ${I^{2}}_{rms} \times R$

$= (7.255)^{2} \times 15 W = 789.5 W .$

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Electrostatic Potential And Capacitance

Suppose the circuit in Question 7.18 has a resistance of 15 Ω Obtain the average power transferred to each element of the circuit, and the total power absorbed.

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