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

7.19 Suppose the circuit in Exercise 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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Payal Gupta | Contributor-Level 10

4 months ago

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

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

Resistance of the resistor, R = 15 Ω

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

Since the elements are connected in series, the impedance is given by

Z = $\sqrt[]{R^{2} + ({ω L - \frac{1}{ω C})}^{2}}$

= $\sqrt[]{15^{2} + ({100 π \times 80 \times 10^{- 3} - \frac{1}{100 π \times 60 \times 10^{- 6}})}^{2}}$

= $\sqrt[]{225 + ({25.133 - 53.052)}^{2}}$

= 31.693 Ω

Current flowing in the circuit, I = $\frac{V}{Z}$ = $\frac{230}{31.693}$ = 7.26 A

Average power transferred to resistance is given as $P_{R}$ = $I^{2}$ R = ${(7.26)}^{2} \times 15$ = 789.97 W

Average power trans

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

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

Resistance of the resistor, R = 15 Ω

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

Since the elements are connected in series, the impedance is given by

Z = $\sqrt[]{R^{2} + ({ω L - \frac{1}{ω C})}^{2}}$

= $\sqrt[]{15^{2} + ({100 π \times 80 \times 10^{- 3} - \frac{1}{100 π \times 60 \times 10^{- 6}})}^{2}}$

= $\sqrt[]{225 + ({25.133 - 53.052)}^{2}}$

= 31.693 Ω

Current flowing in the circuit, I = $\frac{V}{Z}$ = $\frac{230}{31.693}$ = 7.26 A

Average power transferred to resistance is given as $P_{R}$ = $I^{2}$ R = ${(7.26)}^{2} \times 15$ = 789.97 W

Average power transferred to capacitor, $P_{C}$ = 0

Average power transferred to Inductor, $P_{L}$ = 0

Total power absorbed by the circuit = $P_{R} + P_{C} + P_{L} = 789.97 + 0 + 0 = 789.97 W$

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7.19 Suppose the circuit in Exercise 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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