(a) Explanation Formula to calculate the impedance is, Z = R 2 + X C 2 (1) As in above expression, it is seen that that the impedance is directly proportional to the capacitive reactance. Formula for the capacitive reactance is, X C = 1 2 π f C (2) As in the above expression the capacitive reactance is inversely proportional to the frequency. So greater the frequency lesser be the capacitive reactance and lesser the capacitive reactance lesser be the impedance. Lesser the impedance greater be the flow of current in the circuit by the expression given below (b) To determine The best explanation among the given situation.

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ISBN: 8220103026918

Author: Walker

Publisher: PEARSON

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Chapter 24, Problem 73GP

(a)

To determine

Whether the intensity of the light bulb will increase, decrease or stay the same due to the increase in the generator’s frequency.

(b)

To determine

The best explanation among the given situation.

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Chapter 24, Problem 72GP

Chapter 24, Problem 74GP

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charge and the current (i) is zero, then P= 0. If q =0 at time t =0, then =±n/2. Example: A 300-V dc power supply is used to charge a 25-µF capacitor. After the capacitor is fully charged, it is disconnected from the power supply and connected across a 10-mH inductor. The resistance in the circuit is negligible. (a) Find the frequency and period of oscillation of the circuit. (b) Find the capacitor charge and the circuit current 1.2 ms after the inductor and capacitor are connected. Then find for the magnetic and electric energies (c) at and (d) at t = 1.2 ms. Given: C = 25 x 106 F L = 10 x 103 H t = 1.2 x 103 s (c) Solving for magnetic (UB) and electric (UE) energies at time t =0. Solution: (7.5 x 10-3 C)? (a) Solving for angular frequency (w)and period (T) Ug =Li? = 0 Ug = 2C - 1.1 J 2(25 x 10-6 F) 1 = 2.03 x 10° rad/s (10 x 10-3H)(25 x 10¬°F) (d) Solving for magnetic (UB) and electric (UE) energies at time t =1.2 ms. 2. 03 х 103 Рad Ug = }Li² = }(10 × 10-3 H)(-10 A)² = 0.5 J f = 320…

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Chapter 24 Solutions

EBK PHYSICS

Ch. 24.1 - A circuit has a sinusoidally varying current whose...Ch. 24.2 - Circuit 1 has a capacitor with a capacitance C and...Ch. 24.3 - The two circuits shown in Figure 24-14 have...Ch. 24.4 - A circuit consists of an ac generator and an...Ch. 24.5 - Prob. 5EYU Ch. 24.6 - An RLC circuit has a resonance frequency of 60 Hz....Ch. 24 - Why is the current in an ac circuit not always in...Ch. 24 - An LC circuit is driven at a frequency higher than...Ch. 24 - An LC circuit is driven at a frequency lower than...Ch. 24 - Prob. 4CQ

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Whether the intensity of the light bulb will increase, decrease or stay the same due to the increase in the generator’s frequency.

Solution Summary: The author explains that the intensity of the light bulb will increase due to the increase in the generator’s frequency.

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Whether the intensity of the light bulb will increase, decrease or stay the same due to the increase in the generator’s frequency.

The best explanation among the given situation.

Want to see the full answer?

Chapter 24 Solutions