Ask & Answer Question

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

7.15 A 100 μF capacitor in series with a 40 Ω resistance is connected to a 110 V, 60 Hz supply.

(a) What is the maximum current in the circuit?

(b) What is the time lag between the current maximum and the voltage maximum?

3 Views | Posted 4 months ago

Asked by Shiksha User

1 Answer

Answered by

Payal Gupta | Contributor-Level 10

4 months ago

7.15 Capacitance of the capacitor, C = 100 $μ F$ = 100 $\times 10^{- 6}$ F

Resistance of the resistor, R = 40 Ω

Supply voltage, V = 110 V

Frequency of the supply voltage, $ν$ = 60 Hz

Angular frequency, $ω$ = 2 $π ν$ = 2 $π \times 60$ rad/s

For a RC circuit, the impedance Z, is given by Z = $\sqrt[]{R^{2} + \frac{1}{ω^{2} C^{2}}}$

Z = $\sqrt[]{40^{2} + \frac{1}{{(2 π \times 60)}^{2} {\times (100 \times 10^{- 6})}^{2}}}$ = 47.996 Ω

Peak voltage, $V_{0}$ = $\sqrt 2 \times V$ = $\sqrt 2 \times 110$ = 155.56 V

Peak current, $I_{0}$ = $\frac{V_{0}}{Z}$ = $\frac{155.56}{47.996}$ = 3.24 A

In a capacitor circuit, the voltage lags behind the current by a phase angle of $\emptyset$ . This angle is given by the relation

$\tan ? \emptyset$ = $\frac{\frac{1}{ω C}}{R}$ = $\frac{1}{ω C R}$ = $\frac{1}{2 π \times 60 \times 100 \times 10^{- 6} \times 40}$ = 0.663

$\emptyset = 33.55 ° = \frac{33.55 ° \times π}{180}$ &

...more

7.15 Capacitance of the capacitor, C = 100 $μ F$ = 100 $\times 10^{- 6}$ F

Resistance of the resistor, R = 40 Ω

Supply voltage, V = 110 V

Frequency of the supply voltage, $ν$ = 60 Hz

Angular frequency, $ω$ = 2 $π ν$ = 2 $π \times 60$ rad/s

For a RC circuit, the impedance Z, is given by Z = $\sqrt[]{R^{2} + \frac{1}{ω^{2} C^{2}}}$

Z = $\sqrt[]{40^{2} + \frac{1}{{(2 π \times 60)}^{2} {\times (100 \times 10^{- 6})}^{2}}}$ = 47.996 Ω

Peak voltage, $V_{0}$ = $\sqrt 2 \times V$ = $\sqrt 2 \times 110$ = 155.56 V

Peak current, $I_{0}$ = $\frac{V_{0}}{Z}$ = $\frac{155.56}{47.996}$ = 3.24 A

In a capacitor circuit, the voltage lags behind the current by a phase angle of $\emptyset$ . This angle is given by the relation

$\tan ? \emptyset$ = $\frac{\frac{1}{ω C}}{R}$ = $\frac{1}{ω C R}$ = $\frac{1}{2 π \times 60 \times 100 \times 10^{- 6} \times 40}$ = 0.663

$\emptyset = 33.55 ° = \frac{33.55 ° \times π}{180}$ rad

$T i m e l a g, t = \frac{\emptyset}{ω}$ = $\frac{\frac{33.55 ° \times π}{180}}{2 π \times 60}$ = $\frac{33.55 \times π}{2 π \times 60 \times 180}$ = 1.55 $\times 10^{- 3}$ s = 1.55 ms

Hence the time lag between maximum current and maximum voltage is 1.55 ms.

less

0 Upvotes 0 Downvotes

7.15 A 100 μF capacitor in series with a 40 Ω resistance is connected to a 110 V, 60 Hz supply.

(a) What is the maximum current in the circuit?

(b) What is the time lag between the current maximum and the voltage maximum?

1 Answer

Similar Questions for you

Learn more about...

Most viewed information

Share Your College Life Experience

Didn't find the answer you were looking for?

Need guidance on career and education? Ask our experts

Your Question

7.15 A 100 μF capacitor in series with a 40 Ω resistance is connected to a 110 V, 60 Hz supply. (a) What is the maximum current in the circuit? (b) What is the time lag between the current maximum and the voltage maximum?

1 Answer

Similar Questions for you

Learn more about...

Most viewed information

Share Your College Life Experience

Didn't find the answer you were looking for?

Need guidance on career and education? Ask our experts

Your Question

7.15 A 100 μF capacitor in series with a 40 Ω resistance is connected to a 110 V, 60 Hz supply.

(a) What is the maximum current in the circuit?

(b) What is the time lag between the current maximum and the voltage maximum?