The current through a wire depends on time as i = i 0 + α sinπ t , where i 0 = 10 A and α = π 2 A . Find the charge crossed through a section of the wire in 3 seconds , and the average current for that interval.

Given, current as function of time, i = i 0 + α sinπ t , and i 0 = 10 A , α = π 2 As we know that the rate of flow of charge in a conductor is current which is given as, i = d q d t . So we can write d q = i d t , Now to get total charge we need to integrate d q from time interval t = 0 s to t = 3 s , i.e., ∫ d q = ∫ 0 3 i d t ⇒ q = ∫ 0 3 i 0 + α sinπ t d t ⇒ q = ∫ 0 3 i 0 d t + ∫ 0 3 α sinπ t d t ⇒ q = 30 + 1 = 31 . So, the charge crossed through the section of the wire in 3 s = 31 C and the average current for the same interval is, total charge crossed ∆ t ⇒ 31 3 = 10 . 33 A .

The current through a wire depends on time as i = i 0 + α sinπ t , where i 0 = 10 A and α = π 2 A . Find the charge crossed through a section of the wire in 3 seconds , and the average current for that interval.

Given, current as function of time, i = i 0 + α sinπ t , and i 0 = 10 A , α = π 2 As we know that the rate of flow of charge in a conductor is current which is given as, i = d q d t . So we can write d q = i d t , Now to get total charge we need to integrate d q from time interval t = 0 s to t = 3 s , i.e., ∫ d q = ∫ 0 3 i d t ⇒ q = ∫ 0 3 i 0 + α sinπ t d t ⇒ q = ∫ 0 3 i 0 d t + ∫ 0 3 α sinπ t d t ⇒ q = 30 + 1 = 31 . So, the charge crossed through the section of the wire in 3 s = 31 C and the average current for the same interval is, total charge crossed ∆ t ⇒ 31 3 = 10 . 33 A .

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The current through a wire depends on time as $i = i_{0} + α sinπ t$ , where $i_{0} = 10 A$ and $α = \frac{π}{2} A$ . Find the charge crossed through a section of the wire in $3 seconds$ , and the average current for that interval.

Important Questions on Alternating Current

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Alpha Question Bank for Engineering: Physics>Chapter 28 - Alternating Current>Exercise - 2>Q 1

A coil has an inductance of

\frac{2.2}{π} H

and is joined in series with a resistance of

220 Ω .

When an alternating e.m.f. of

220 V

50

c.p.s. is applied to it, then the watt-less component of the RMS current in the circuit is

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Alpha Question Bank for Engineering: Physics>Chapter 28 - Alternating Current>Exercise - 2>Q 2

The current in a circuit containing a capacitance

C

and a resistance

R

in series leads over the applied voltage of frequency

\frac{ω}{2 π}

by,

HARD

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Alpha Question Bank for Engineering: Physics>Chapter 28 - Alternating Current>Exercise - 2>Q 3

An inductor $(L = \frac{1}{100 π} H)$ , a capacitor $(C = \frac{1}{500 π} F)$ and a resistance $(3 Ω)$ are connected in series with an AC voltage source as shown in the figure. The voltage of the AC source is given as $V = 10 \cos (100 πt) volt$ . What will be the potential difference between $A$ and $B$ ?

Question Image

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Alpha Question Bank for Engineering: Physics>Chapter 28 - Alternating Current>Exercise - 2>Q 4

An ac voltage source, $V = V_{0} sinω t$ is connected across resistance $R$ and capacitance $C$ as shown in figure. It is given that, $R = \frac{1}{ω C}$ . The peak current is $I_{0}$ . If the angular frequency of the voltage source is changed to $\frac{ω}{\sqrt{3}}$ keeping $R$ and $C$ fixed, then the new peak current in the circuit is,

Question Image

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Alpha Question Bank for Engineering: Physics>Chapter 28 - Alternating Current>Exercise - 2>Q 5

For a LCR series circuit with an A.C. source of angular frequency

ω .

HARD

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Alpha Question Bank for Engineering: Physics>Chapter 28 - Alternating Current>Exercise - 2>Q 6

L C R

series circuit with

100 Ω

resistance is connected to an

A C

source of

200 V

and angular frequency

300 radians per second

. When only the capacitance is removed, the current lags behind the voltage by

60 °

. When only the inductance is removed, the current leads the voltage by

60 °

. Then, the current and power dissipated in

L C R

circuit are, respectively

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Alpha Question Bank for Engineering: Physics>Chapter 28 - Alternating Current>Exercise - 2>Q 7

In an

L R

series circuit

(L = \frac{175}{11} mH and R = 12 Ω)

, a variable emf source

(V = V_{0} sinω t) of V_{rms} =130 \sqrt{2} V

and frequency

50 Hz

is applied. The current amplitude in the circuit and phase of current with respect to voltage are, respectively

(use, π = \frac{22}{7})

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Alpha Question Bank for Engineering: Physics>Chapter 28 - Alternating Current>Exercise - 2>Q 8

In LCR circuit at resonance, the current in the circuit is

10 \sqrt{2} A

. If now the frequency of the source is changed such that now the current lags by

45 °

than applied voltage in the circuit, which of the following is correct?

The current through a wire depends on time as i=i0+αsinπt, where i0=10 A and α=π2 A. Find the charge crossed through a section of the wire in 3 seconds, and the average current for that interval.

Important Questions on Alternating Current

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The current through a wire depends on time as $i = i_{0} + α sinπ t$ , where $i_{0} = 10 A$ and $α = \frac{π}{2} A$ . Find the charge crossed through a section of the wire in $3 seconds$ , and the average current for that interval.