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

7.20 A series LCR circuit with L = 0.12 H, C = 480 nF, R = 23 Ω is connected to a 230 V variable frequency supply.

(a) What is the source frequency for which current amplitude is maximum. Obtain this maximum value.

(b) What is the source frequency for which average power absorbed by the circuit is maximum. Obtain the value of this maximum power.

(c) For which frequencies of the source is the power transferred to the circuit half the power at resonant frequency? What is the current amplitude at these frequencies?

(d) What is the Q-factor of the given circuit?

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Payal Gupta | Contributor-Level 10

4 months ago

7.20 Inductance, L = 0.12 H

Capacitance, C = 480 nF = 480 $\times 10^{- 9}$ F

Resistance, R = 23 Ω

Supply voltage, V = 230 V

Peak voltage is given as, $V_{0}$ = $\sqrt[]{2} \times 230$ = 325.27 V

Current flowing in the circuit is given by the relation,

$I_{0} = \frac{V_{0}}{\sqrt[]{R^{2} + ({ω L - \frac{1}{ω C})}^{2}}}$ where $I_{0}$ = maximum at resonance

$A t r e s o n a n c e w e h a v e ω_{R} L - \frac{1}{ω_{R} C} = 0$ , where $ω_{R}$ = Resonance angular frequency.

Hence, $ω_{R}$ = $\frac{1}{\sqrt[]{L C}}$ = $\frac{1}{\sqrt[]{0.12 \times 480 \times 10^{- 9}}}$ = 4166.67 rad/s

Therefore, resonating frequency $ν_{R}$ = $\frac{ω_{R}}{2 π}$ = 663.15 Hz

Maximum current ${I_{0}}_{m a x}$ = $\frac{V_{0}}{R}$ = $\frac{325.27}{23}$ = 14.14 A

Maximum average power absorbed by the circuit is given as

${P_{a v g}}_{m a x}$ = $\frac{1}{2} {I_{0}}_{m a x}^{2}$ R = $\frac{1}{2} \times {14.14}^{2} \times 23$ = 2300 W

The power transferre

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7.20 Inductance, L = 0.12 H

Capacitance, C = 480 nF = 480 $\times 10^{- 9}$ F

Resistance, R = 23 Ω

Supply voltage, V = 230 V

Peak voltage is given as, $V_{0}$ = $\sqrt[]{2} \times 230$ = 325.27 V

Current flowing in the circuit is given by the relation,

$I_{0} = \frac{V_{0}}{\sqrt[]{R^{2} + ({ω L - \frac{1}{ω C})}^{2}}}$ where $I_{0}$ = maximum at resonance

$A t r e s o n a n c e w e h a v e ω_{R} L - \frac{1}{ω_{R} C} = 0$ , where $ω_{R}$ = Resonance angular frequency.

Hence, $ω_{R}$ = $\frac{1}{\sqrt[]{L C}}$ = $\frac{1}{\sqrt[]{0.12 \times 480 \times 10^{- 9}}}$ = 4166.67 rad/s

Therefore, resonating frequency $ν_{R}$ = $\frac{ω_{R}}{2 π}$ = 663.15 Hz

Maximum current ${I_{0}}_{m a x}$ = $\frac{V_{0}}{R}$ = $\frac{325.27}{23}$ = 14.14 A

Maximum average power absorbed by the circuit is given as

${P_{a v g}}_{m a x}$ = $\frac{1}{2} {I_{0}}_{m a x}^{2}$ R = $\frac{1}{2} \times {14.14}^{2} \times 23$ = 2300 W

The power transferred to the circuit is half the power at resonant frequency

Frequencies at which power transferred is half = $ω_{R} \pm$ Δ $ω$ = 2 $π (ν_{R} \pm Δ ν)$ where,

Δ $ω = \frac{R}{2 L}$ = $\frac{23}{2 \times 0.12}$ = 95.83 rad/s

$Δ ν$ = $\frac{Δ ω}{2 π}$ = $\frac{95.83}{2 \times π}$ = 15.25 Hz

Therefore $ν_{R} \pm Δ ν$ = 663.15 $\pm$ 15.25 = 647.9 Hz and 678.4 Hz

Hence, at 647.9 Hz and 678.4 Hz frequencies, the power transferred is half.

At these frequencies, current amplitude can be given as:

I’ = $\frac{1}{\sqrt 2} \times {I_{0}}_{m a x}$ = $\frac{14.14}{\sqrt 2}$ = 10 A

Q factor of the given circuit can be obtained using the relation,

Q = $\frac{ω_{R} L}{R}$ = $\frac{4166.67 \times 0.12}{23}$ = 21.74

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