An alternating voltage $mathrmV(t) = 220sin 100pi mathrmt$ volt is applied to a purely resistive load of $50Omega$ $50\Omega$. The time taken for the current to rise from half of the peak value to the peak value is:

A. $5 mathrm~ms$ $5 \mathrm{~ms}$
B. $3.3 mathrm~ms$ $3.3 \mathrm{~ms}$
C. $7.2 mathrm~ms$ $7.2 \mathrm{~ms}$
D. $2.2 mathrm~ms$ $2.2 \mathrm{~ms}$

Solution & Explanation

### Related Formula

I(t) = I_0 sin(omega t)

omega = frac2piT

### Core Logic Since the load is purely resistive, the current is in phase with the voltage:

I(t) = I_0 sin(100pi t)

. We need the time difference between the instant current reaches

fracI_02

and the instant it reaches

I_0

. ### Step 1: Time for Half Peak

I(t_1) = fracI_02 implies I_0 sin(omega t_1) = fracI_02

sin(omega t_1) = frac12 implies omega t_1 = fracpi6

### Step 2: Time for Peak

I(t_2) = I_0 implies I_0 sin(omega t_2) = I_0

sin(omega t_2) = 1 implies omega t_2 = fracpi2

### Step 3: Calculate Time Interval The required time interval is

Delta t = t_2 - t_1

omega Delta t = fracpi2 - fracpi6 = fracpi3

Delta t = fracpi3omega

Given

omega = 100pi

Delta t = fracpi3 times 100pi = frac1300 mathrm~s = 3.33 mathrm~ms

### Pattern Recognition In an AC sine wave, going from 0 to peak takes

T/4

. Going from 0 to half peak takes

T/12

. Therefore, going from half peak to peak takes

T/4 - T/12 = T/6

. Calculate

T/6

directly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Alternating Current

Reference Study Guides

Alternating Current

More Alternating Current Previous-Year Questions — Page 2

Q22 jee_main_2025_07_april_evening AC Circuits with LCR

An inductor of reactance

100Omega

, a capacitor of reactance

50Omega

, and a resistor of resistance

50Omega

are connected in series with an AC source of

10mathrm~V

50mathrm~Hz

. Average power dissipated by the circuit is ______ W. [cite: 185, 186]

Numerical Answer. Answer: 1 to 1

Solution

### Related Formula

Z = sqrtR^2 + (X_L - X_C)^2

[cite: 786]

P = I_textrms^2 R = fracV_textrms^2 RZ^2

[cite: 785, 786] ### Core Logic First, find the total impedance

Z

of the LCR circuit: [cite: 185, 786]

Z = sqrt50^2 + (100 - 50)^2 = sqrt50^2 + 50^2 = 50sqrt2\ Omega

[cite: 185, 786] Now calculate the average power dissipation: [cite: 185, 786]

P = frac(10)^2 times 50(50sqrt2)^2 = frac100 times 502500 times 2 = frac50005000 = 1\ textW

[cite: 787] ### Pattern Recognition Remember that only the resistive element dissipates real thermal power over a complete cycle[cite: 784, 785]. Reactive elements like ideal inductors and capacitors store and release energy alternately without net consumption. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Alternating Current

Q16 jee_main_2025_24_jan_morning RMS Value of Alternating Current

An alternating current is given by

I=I_Atext sin omega t +I_Btext cos omega t.

The r.m.s. current will be :-

A. $sqrtI_A^2+I_B^2$ $\sqrt{I_{A}^{2}+I_{B}^{2}}$
B. $fracsqrtI_A^2+I_B^22$ $\frac{\sqrt{I_{A}^{2}+I_{B}^{2}}}{2}$
C. $sqrtfracI_A^2+I_B^22$ $\sqrt{\frac{I_{A}^{2}+I_{B}^{2}}{2}}$
D. $frac|I_A+I_B|sqrt2$ $\frac{|I_{A}+I_{B}|}{\sqrt{2}}$

Solution

### Related Formula The root-mean-square current value

I_textrms

for a periodic function is defined as:

I_textrms = sqrtfrac1Tint_0^T I^2 dt

For a single sinusoidal term

I = I_0sin(omega t + phi)

, the root-mean-square value simplifies directly to:

I_textrms = fracI_0sqrt2

### Core Logic We can combine the orthogonal sine and cosine components into a single phase-shifted wave:

I = I_Asinomega t + I_Bcosomega t = sqrtI_A^2 + I_B^2sin(omega t + phi)

where peak current amplitude corresponds to:

I_0 = sqrtI_A^2 + I_B^2

### Step 1: Calculating RMS Value Using the standard peak-to-RMS conversion factor:

I_textrms = fracI_0sqrt2 = fracsqrtI_A^2 + I_B^2sqrt2 = sqrtfracI_A^2 + I_B^22

### Pattern Recognition Orthogonal sine and cosine functions are independent. Their mean squared averages add linearly, so you can think of it as a vector addition:

fracI_A^22 + fracI_B^22

under the square root. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Alternating Current

Q1 jee_main_2025_29_jan_morning Choke Coil and AC Circuits

Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A): Choke coil is simply a coil having a large inductance but a small resistance. Choke coils are used with fluorescent mercury-tube fittings. If household electric power is directly connected to a mercury tube, the tube will be damaged. Reason (R): By using the choke coil, the voltage across the tube is reduced by a factor

left(mathrmR / sqrtmathrmR^2 + omega^2mathrmL^2right)

, where

omega

is frequency of the supply across resistor R and inductor L. If the choke coil were not used, the voltage across the resistor would be the same as the applied voltage. In the light of the above statements, choose the most appropriate answer from the options given below:

A. $textBoth (A) and (R) are true but (R) is not the correct explanation of (A).$
B. $text(A) is false but (R) is true.$ $\text{(A) is false but (R) is true.}$
C. $textBoth (A) and (R) are true and (R) is the correct explanation of (A).$
D. $text(A) is true but (R) is false.$ $\text{(A) is true but (R) is false.}$

Solution

### Related Formula

Z = sqrtR^2 + (omega L)^2

cos phi = fracRZ

### Core Logic A choke coil has a high inductance

L

and low resistance

R

. It reduces the current through the mercury tube without wasting electrical power as heat. Statement (A) is correct. The average power dissipation is controlled by the low power factor

cos phi

. Reason (R) correctly explains that without a choke coil, the direct supply voltage could cause an overflow of current, destroying the tube filament. ### Pattern Recognition Choke coils use high

L

and low

R

to minimize power loss while reducing circuit current efficiently. ### Chapter Mix Class 12 Physics: Alternating Current

Q43 jee_main_2024_01_february_morning LCR Resonance

In a series LCR circuit, the capacitance is changed from

C

4C

. To keep the resonance frequency unchanged, the new inductance should be:

A. reduced by $frac14 L$ $\frac{1}{4} L$
B. increased by $2L$ $2L$
C. reduced by $frac34 L$ $\frac{3}{4} L$
D. increased to $4L$ $4L$

Solution

### Related Formula Resonant angular frequency formula:

omega = frac1sqrtLC

### Core Logic For

omega' = omega

frac1sqrtL'C' = frac1sqrtLC implies L'C' = LC

Given new capacitance

C' = 4C

L'(4C) = LC implies L' = fracL4

### Step 1: Calculate the Change Needed The question asks for the reduction increment or statement matching the dynamic shift:

Delta L = L - L' = L - fracL4 = frac34L

Therefore, the inductance must be reduced by

frac34L

. ### Pattern Recognition Read options carefully. The new value is

frac14L

, which means it must be **reduced by**

frac34L

. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Alternating Current

Q34 jee_main_2024_29_january_evening Power in AC Circuits

In an a.c. circuit, voltage and current are given by:

V = 100sin(100t)text V

and

I = 100sinleft(100t + fracpi3right)text mA

respectively. The average power dissipated in one cycle is:

A. $5text W$ $5\text{ W}$
B. $10text W$ $10\text{ W}$
C. $2.5text W$ $2.5\text{ W}$
D. $25text W$ $25\text{ W}$

Solution

### Related Formula The average power dissipated in an AC circuit is:

P_textavg = V_textrms I_textrms cos(Delta phi)

where: *

V_textrms = fracV_0sqrt2

I_textrms = fracI_0sqrt2

Delta phi

is the phase difference between voltage and current. ### Core Logic From the given equations: * Peak Voltage,

V_0 = 100text V

* Peak Current,

I_0 = 100text mA = 100 times 10^-3text A = 0.1text A

* Phase Difference,

Delta phi = fracpi3

### Step 1: Calculate Average Power Substitute these values into the average power formula:

P_textavg = left( frac100sqrt2 right) times left( frac100 times 10^-3sqrt2 right) times cosleft( fracpi3 right)

P_textavg = frac10^4 times 10^-32 times frac12

P_textavg = frac104 = 2.5text W

### Pattern Recognition Remember to look closely at units! The current is given in mA (

10^-3text A

). Missing this conversion leads directly to the trap answer (

2500text W

or similar scaling errors). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Alternating Current

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An alternating voltage $mathrmV(t) = 220sin 100pi mathrmt$ volt is applied to a purely resistive load of $50Omega$ $50\Omega$. The time taken for the current to rise from half of the peak value to the peak value is:

Solution & Explanation

Reference Study Guides

More Alternating Current Previous-Year Questions — Page 2

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An alternating voltage mathrmV(t) = 220sin 100pi mathrmt$\mathrm{V(t)} = 220\sin 100\pi \mathrm{t}$ volt is applied to a purely resistive load of 50Omega$50\Omega$. The time taken for the current to rise from half of the peak value to the peak value is:

Solution & Explanation

Reference Study Guides

More Alternating Current Previous-Year Questions — Page 2

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An alternating voltage $mathrmV(t) = 220sin 100pi mathrmt$ volt is applied to a purely resistive load of $50Omega$ $50\Omega$. The time taken for the current to rise from half of the peak value to the peak value is: