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

7.11 Figure 7.21 shows a series LCR circuit connected to a variable frequency 230 V source. L = 5.0 H, C = 80μF, R = 40 Ω.

(a) Determine the source frequency which drives the circuit in resonance.

(b) Obtain the impedance of the circuit and the amplitude of current at the resonating frequency.

(c) Determine the rms potential drops across the three elements of the circuit. Show that the potential drop across the LC combination is zero at the resonating frequency.

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Payal Gupta | Contributor-Level 10

4 months ago

7.11 Inductance of the Inductor, L = 5.0 H

Resistance of the resistor, R = 40 Ω

Capacitance of the capacitor, C = 80 $μ F =$ 80 $\times 10^{- 6}$ F

Potential of the voltage source, V = 230 V

Resonance angular frequency is given as

$ω_{r}$ = $\frac{1}{\sqrt[]{L C}}$ = $\frac{1}{\sqrt[]{5 \times 80 \times 10^{- 6}}}$ = 50 rad/s

$H e n c e, t h e c i r c u i t w i l l c o m e i n r e s o n a n c e f o r a s o u r c e f r e q u e n c y o f 50 r a d / s$

The impedance of the circuit is given as

Z = $\sqrt[]{R^{2} + {(X_{L -} X_{C})}^{2}}$ where $X_{L}$ = Inductive reactance and $X_{C}$ = Capacitive reactance

At resonance, $X_{L}$ = $X_{C}$ so Z = R = 40Ω

Amplitude of the current at the resonating frequency is given as

$I_{o}$ = $\frac{V_{0}}{Z}$ where $V_{0}$ = Peak voltage = $\sqrt 2$ V

$I_{o}$ = $\frac{\sqrt 2 V}{Z}$ = $\frac{\sqrt 2 \times 230}{40}$ = 8.13 A

Hence, at reson

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7.11 Inductance of the Inductor, L = 5.0 H

Resistance of the resistor, R = 40 Ω

Capacitance of the capacitor, C = 80 $μ F =$ 80 $\times 10^{- 6}$ F

Potential of the voltage source, V = 230 V

Resonance angular frequency is given as

$ω_{r}$ = $\frac{1}{\sqrt[]{L C}}$ = $\frac{1}{\sqrt[]{5 \times 80 \times 10^{- 6}}}$ = 50 rad/s

$H e n c e, t h e c i r c u i t w i l l c o m e i n r e s o n a n c e f o r a s o u r c e f r e q u e n c y o f 50 r a d / s$

The impedance of the circuit is given as

Z = $\sqrt[]{R^{2} + {(X_{L -} X_{C})}^{2}}$ where $X_{L}$ = Inductive reactance and $X_{C}$ = Capacitive reactance

At resonance, $X_{L}$ = $X_{C}$ so Z = R = 40Ω

Amplitude of the current at the resonating frequency is given as

$I_{o}$ = $\frac{V_{0}}{Z}$ where $V_{0}$ = Peak voltage = $\sqrt 2$ V

$I_{o}$ = $\frac{\sqrt 2 V}{Z}$ = $\frac{\sqrt 2 \times 230}{40}$ = 8.13 A

Hence, at resonance, the impedance of the circuit is 40 Ω and the amplitude of the current is 8.13 A.

rms potential drop across the inductor is given by

${{(V}_{l})}_{r m s}$ = $I_{r m s}$ $\times ω_{r} L$

$I_{r m s} =$ $\frac{I_{o}}{\sqrt 2}$ = $\frac{8.13}{\sqrt 2}$ = 5.75 A

$H e n c e,$ ${{(V}_{l})}_{r m s}$ = $5.75$ $\times 50 \times 5$ = 1437.5 V

Potential drop across capacitor,

${{(V}_{C})}_{r m s}$ = $I_{r m s}$ $\times \frac{1}{ω_{r} \times C}$

$H e n c e,$ ${{(V}_{l})}_{r m s}$ = $5.75$ $\times \frac{1}{50 \times 80 \times 10^{- 6}}$ = 1437.5 V

Potential drop across resistor,

${{(V}_{R})}_{r m s}$ = $I_{r m s}$ $\times R$ = 5.75 $\times 40$ = 230 V

Potential drop across LC combination

${{(V}_{L C})}_{r m s}$ = $I_{r m s}$ $(X_{L} - X_{C})$

At resonance, $X_{L}$ = $X_{C})$ , hence

${{(V}_{L C})}_{r m s} = 0$

Hence, it is proved that the potential drop across the LC combination is zero at resonating frequency.

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