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Alternating Current Physics Ncert Solutions Class 12th Physics Class 12th Alternating Current Overview

7.18 A circuit containing a 80 mH inductor and a 60 $μ$ F capacitor in series is connected to a 230 V, 50 Hz supply. The resistance of the circuit is negligible.

(a) Obtain the current amplitude and rms values.

(b) Obtain the rms values of potential drops across each element.

(c) What is the average power transferred to the inductor?

(d) What is the average power transferred to the capacitor?

(e) What is the total average power absorbed by the circuit? [‘Average’ implies ‘averaged over one cycle’.]

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Payal Gupta | Contributor-Level 10

4 months ago

7.18 Inductance, L = 80 mH = 80 $\times 10^{- 3}$ H

Capacitance, C = 60 $μ F$ = 60 $\times 10^{- 6}$ F

Supply voltage, V = 230 V

Supply frequency, $ν$ = 50 Hz

Peak voltage, $V_{0}$ = V $\sqrt 2$ = 325.27 V

Angular frequency, $ω$ = 2 $π ν$ = 2 $π \times ν$ = 2 $π \times 50$ = 100 $π$ rad/s

Maximum current is given as:

$I_{0}$ = $\frac{V_{0}}{(ω L - \frac{1}{ω C})}$ = $\frac{325.27}{(100 π \times 80 \times 10^{- 3} - \frac{1}{100 π \times 60 \times 10^{- 6}})}$ = $-$ 11.65 A

The negative sign is due to $ω L$ $< \frac{1}{ω C}$

Amplitude of maximum current $|I_{0}|$ = 11.65 A

rms value of the current, I = $\frac{I_{0}}{\sqrt 2}$ = $\frac{- 11.65}{\sqrt 2}$ = -8.24 A

(i) Potential difference across inductor, $V_{L}$ = I $\times ω L$ = 8.24 $\times 100 π \times$ 80 $\times 10^{- 3}$ V = 207.09 V

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7.18 Inductance, L = 80 mH = 80 $\times 10^{- 3}$ H

Capacitance, C = 60 $μ F$ = 60 $\times 10^{- 6}$ F

Supply voltage, V = 230 V

Supply frequency, $ν$ = 50 Hz

Peak voltage, $V_{0}$ = V $\sqrt 2$ = 325.27 V

Angular frequency, $ω$ = 2 $π ν$ = 2 $π \times ν$ = 2 $π \times 50$ = 100 $π$ rad/s

Maximum current is given as:

$I_{0}$ = $\frac{V_{0}}{(ω L - \frac{1}{ω C})}$ = $\frac{325.27}{(100 π \times 80 \times 10^{- 3} - \frac{1}{100 π \times 60 \times 10^{- 6}})}$ = $-$ 11.65 A

The negative sign is due to $ω L$ $< \frac{1}{ω C}$

Amplitude of maximum current $|I_{0}|$ = 11.65 A

rms value of the current, I = $\frac{I_{0}}{\sqrt 2}$ = $\frac{- 11.65}{\sqrt 2}$ = -8.24 A

(i) Potential difference across inductor, $V_{L}$ = I $\times ω L$ = 8.24 $\times 100 π \times$ 80 $\times 10^{- 3}$ V = 207.09 V

(ii) Potential drop across capacitor, $V_{C}$ = I $\times \frac{1}{ω C}$ = 8.24 $\times \frac{1}{100 π \times 60 \times 10^{- 6}}$ = 437.14 V

Average power consumed by the inductor is zero as actual voltage leads the current by $\frac{π}{2}$

Average power consumed by the capacitor is zero as the voltage lags current by $\frac{π}{2}$

The total power absorbed ( average over one cycle) is zero.

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