Physics Crash Course (Based on Revised Syllabus - 2023)>Chapter 7 - Alternating Current>Exercise>Q 13

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As the frequency of an AC circuit increases, the current first increases and then decreases. What combination of circuit element is most likely to comprise the circuit?

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Important Questions on Alternating Current

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Physics Crash Course (Based on Revised Syllabus - 2023)>Chapter 7 - Alternating Current>Exercise>Q 14

In the Figure given below, the series LCR circuit is shown which is connected to a source of variable frequency and a voltage of $230 V$ . The values of the inductor, capacitor and resistor are given below:

$L = 5.0 H, C = 80 μF, R = 40 Ω$

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Derive the impedance of the network and the amplitude of current at the frequency obtained at resonance.

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Physics Crash Course (Based on Revised Syllabus - 2023)>Chapter 7 - Alternating Current>Exercise>Q 15

The voltage over a cycle varies as

V = \{\begin{cases} V_{0} \sin ω t for 0 \leq t \leq \frac{π}{ω} \\ - V_{0} \sin ω t for \frac{π}{ω} \leq t \leq \frac{2 π}{ω} \end{cases}

The average value of the voltage for one cycle is

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Physics Crash Course (Based on Revised Syllabus - 2023)>Chapter 7 - Alternating Current>Exercise>Q 16

If a capacitor of

8 μF

is connected to a

220 V, 100 H z

, AC source and the current passing through it is

65 m A

then the rms voltage across it is

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Physics Crash Course (Based on Revised Syllabus - 2023)>Chapter 7 - Alternating Current>Exercise>Q 17

In an electrical circuit,

R, L, C

and AC voltage source are all connected in series. When

L

is removed from the circuit, the phase difference between the voltage and the current in the circuit is

\frac{π}{3}

. If instead,

C

is removed from the circuit, the phase difference is again

\frac{π}{3}

. The power factor of the circuit is

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Physics Crash Course (Based on Revised Syllabus - 2023)>Chapter 7 - Alternating Current>Exercise>Q 18

When an

AC

voltage of

220 V

is applied to the capacitor

C

, then

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Physics Crash Course (Based on Revised Syllabus - 2023)>Chapter 7 - Alternating Current>Exercise>Q 19

The power factor of the circuit as shown in figure is:
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Physics Crash Course (Based on Revised Syllabus - 2023)>Chapter 7 - Alternating Current>Exercise>Q 20

In an circuit,

V

and

I

are given by

V = 150 \sin (150 t) V

and

I = 150 \sin (150 t + \frac{π}{3}) A

.
The power dissipated in the circuit is

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Physics Crash Course (Based on Revised Syllabus - 2023)>Chapter 7 - Alternating Current>Exercise>Q 21

In the circuit shown in the figure, if the value of RMS current is $2 A$ , then the power factor of the box is

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