Primary coil of a transformer is connected to 220V ac. Primary and secondary turns of the transforms are 100 and 10…

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Q jee_main_2025_24_jan_morning RMS Value of Alternating Current

An alternating current is given by

I = IA ω t + IB ω t

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

A. $IA²+IB²$ $\sqrt{I_{A}^{2}+I_{B}^{2}}$
B. $IA²+IB²2$ $\frac{\sqrt{I_{A}^{2}+I_{B}^{2}}}{2}$
C. $IA²+IB²2$ $\sqrt{\frac{I_{A}^{2}+I_{B}^{2}}{2}}$
D. $|IA+IB|\sqrt(2)$ $\frac{|I_{A}+I_{B}|}{\sqrt{2}}$

Solution

Related Formula

The root-mean-square current value $Irms$ $I_{\text{rms}}$ for a periodic function is defined as:

Irms = (1)/(T)\int₀T I² dt

For a single sinusoidal term $I = I₀ (ω t + φ)$ $I = I_{0}\sin(\omega t + \phi)$, the root-mean-square value simplifies directly to:

Irms = I₀\sqrt(2)

Core Logic

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

I = IA ω t + IB ω t = IA² + IB² (ω t + φ)

where peak current amplitude corresponds to:

I₀ = IA² + IB²

Step 1: Calculating RMS Value

Using the standard peak-to-RMS conversion factor:

Irms = I₀\sqrt(2) = IA² + IB²\sqrt(2) = IA² + IB²2

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: $(IA²)/(2) + (IB²)/(2)$ $\frac{I_A^2}{2} + \frac{I_B^2}{2}$ under the square root.

Chapter Mix

Class 12 Physics: Alternating Current

Q 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

(R / R² + ω²L²)

, where

ω

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

Solution

Related Formula

Z = \sqrt(R² + (ω L)²)

φ = (R)/(Z)

Core Logic

A choke coil has a high inductance $L$ $L$ and low resistance $R$ $R$. 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 $φ$ $\cos \phi$. 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 $L$ $L$ and low $R$ $R$ 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

to

4C

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

A. reduced by $(1)/(4) L$ $\frac{1}{4} L$
B. increased by $2L$ $2L$
C. reduced by $(3)/(4) L$ $\frac{3}{4} L$
D. increased to $4L$ $4L$

Solution

Related Formula

Resonant angular frequency formula:

ω = 1\sqrt(LC)

Core Logic

For $ω' = ω$ $\omega' = \omega$:

1\sqrt(L'C') = 1\sqrt(LC) L'C' = LC

Given new capacitance $C' = 4C$ $C' = 4C$:

L'(4C) = LC L' = (L)/(4)

Step 1: Calculate the Change Needed

The question asks for the reduction increment or statement matching the dynamic shift:

Δ L = L - L' = L - (L)/(4) = (3)/(4)L

Therefore, the inductance must be reduced by $(3)/(4)L$ $\frac{3}{4}L$.

Pattern Recognition

Read options carefully. The new value is $(1)/(4)L$ $\frac{1}{4}L$, which means it must be reduced by $(3)/(4)L$ $\frac{3}{4}L$.

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 = 100 (100t) V

and

I = 100 (100t + (π)/(3)) mA

respectively. The average power dissipated in one cycle is:

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

Solution

Related Formula

The average power dissipated in an AC circuit is:

Pavg = Vrms Irms (Δ φ)

where:

$Vrms = V₀\sqrt(2)$ $V_{\text{rms}} = \frac{V_0}{\sqrt{2}}$
$Irms = I₀\sqrt(2)$ $I_{\text{rms}} = \frac{I_0}{\sqrt{2}}$
$Δ φ$ $\Delta \phi$ is the phase difference between voltage and current.

Core Logic

From the given equations:

Peak Voltage, $V₀ = 100 V$ $V_0 = 100\text{ V}$
Peak Current, $I₀ = 100 mA = 100 \times 10⁻³ A = 0.1 A$
Phase Difference, $Δ φ = (π)/(3)$ $\Delta \phi = \frac{\pi}{3}$

Step 1: Calculate Average Power

Substitute these values into the average power formula:

Pavg = ( 100\sqrt(2) ) \times ( 100 \times 10⁻³\sqrt(2) ) \times ( (π)/(3) )

Pavg = 10⁴ \times 10⁻³2 \times (1)/(2)

Pavg = (10)/(4) = 2.5 W

Pattern Recognition

Remember to look closely at units! The current is given in mA ( $10⁻³ A$ $10^{-3}\text{ A}$). Missing this conversion leads directly to the trap answer ( $2500 W$ $2500\text{ W}$ or similar scaling errors).

Chapter Mix

Class 12 Physics: Alternating Current

Q48 jee_main_2024_29_jan_morning LC Oscillations

A capacitor of capacitance

100 μ F

is charged to a potential of

12 ~V

and connected to a

6.4 ~mH

inductor to produce oscillations. The maximum current in the circuit would be:

A. 3.2 A
B. 1.5 A
C. 2.0 A
D. 1.2 A

Solution

Related Formula

By conservation of energy in an ideal $LC$ $LC$ oscillating circuit, the maximum electrostatic energy stored in the capacitor equals the maximum magnetic energy stored in the inductor:

(1)/(2) C V² = (1)/(2) L I²

Core Logic

Rearranging the energy equation to express maximum current:

I = V \sqrt((C)/(L))

Given values:

C = 100 μ F = 100 \times 10⁻⁶ ~F

V = 12 ~V

L = 6.4 ~mH = 6.4 \times 10⁻³ ~H

Step 1: Compute Maximum Current

Substituting values:

I = 12 \times 100 \times 10⁻⁶6.4 \times 10⁻³

I = 12 \times 10⁻⁴6.4 \times 10⁻³ = 12 \times \sqrt((1)/(64)) = 12 \times (1)/(8)

I = (12)/(8) = 1.5 ~A

Therefore, the maximum current in the circuit is $1.5 ~A$ $1.5 \mathrm{~A}$.

Pattern Recognition

This problem represents basic harmonic energy transfer between potential states. You can also derive this via peak relations: $I = q₀ ω = (C V) 1\sqrt(L C) = V \sqrt((C)/(L))$ , which bypasses square root conversion hazards if performed methodically.

Chapter Mix

Class 12 Physics: Alternating Current

Solution & Explanation

Related Formula

Core Logic

Step 1: Calculate Secondary Voltage

Step 2: Output Voltage Calculation

Pattern Recognition

Chapter Mix

Reference Study Guides

More Alternating Current Previous-Year Questions — Page 3

Solution

Related Formula

Core Logic

Step 1: Calculating RMS Value

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate the Change Needed

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate Average Power

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Compute Maximum Current

Pattern Recognition

Chapter Mix

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