Solved Numerical Problems on alternating current important questions Alternating Current, LC Oscillations, LCR Circuits, Resonance and Transformers
Alternating Current Numericals • CBSE • NEET • JEE

Solved Numerical Problems on Alternating Current R,L,C,RL,RC,LC,RLC Oscillations, LCR Circuits, Resonance and Transformers

Complete step-by-step solutions for 26 exam questions with derivations, formula cards, SVG diagrams, collapsible answers and Kumar Sir's exam tips.

Section 1: Basic AC Circuits

Voltage and Current Values

Vrms = V0/√2Irms = I0/√2Vavg = 2V0/π

Reactance

XL = ωL = 2πfLXC = 1/ωCω = 2πf

Power Factor

P = VrmsIrmscosφPure L or C: Pavg = 0Pure R: P = V²/R

Question 1

A 100 Ω resistor is connected to a 220 V, 50 Hz AC supply.

CBSE BoardEasyPure Resistor

Given: R = 100 Ω, Vrms = 220 V, f = 50 Hz.

Formula: Irms = Vrms/R, P = V²/R

  1. Irms = 220/100 = 2.2 A.
  2. I0 = √2 Irms = 3.11 A.
  3. In a pure resistor, voltage and current are in phase, so power factor = 1.
  4. P = 220²/100 = 484 W.
Final Answer: Irms = 2.2 A, peak current = 3.11 A, power = 484 W, power factor = 1.

Question 2

The peak voltage of an AC supply is 300 V. What is the RMS voltage?

NEET LevelEasyRMS Value

Given: V0 = 300 V.

Formula: Vrms = V0/√2

  1. Vrms = 300/√2.
  2. Vrms = 212.1 V.
Final Answer: Vrms = 212 V approximately.

Question 3

A 44 mH inductor is connected to a 220 V, 50 Hz AC supply. Determine the RMS value of the current.

NEET LevelMediumPure Inductor

Given: L = 44 mH = 0.044 H, V = 220 V, f = 50 Hz.

Formula: XL = 2πfL, I = V/XL

  1. XL = 2π × 50 × 0.044 = 13.82 Ω.
  2. Irms = 220/13.82 = 15.92 A.
  3. Current lags voltage by 90° in a pure inductor.
Final Answer: Irms = 15.9 A.

Question 4

A 60 μF capacitor is connected to a 110 V, 60 Hz AC supply. Determine the RMS value of the current.

NEET LevelMediumPure Capacitor

Given: C = 60 μF = 60 × 10-6 F, V = 110 V, f = 60 Hz.

Formula: XC = 1/(2πfC), I = V/XC

  1. XC = 1/(2π × 60 × 60 × 10-6) = 44.21 Ω.
  2. Irms = 110/44.21 = 2.49 A.
  3. Current leads voltage by 90° in a pure capacitor.
Final Answer: Irms = 2.49 A.

Question 5

For Exercises 3 and 4, find the net power absorbed by each circuit over a complete cycle.

CBSE BoardEasyWattless Current

Formula: Pavg = VrmsIrmscosφ

  1. For pure inductor, φ = 90°, so cosφ = 0.
  2. PL = VI cos90° = 0 W.
  3. For pure capacitor, φ = 90°, so cosφ = 0.
  4. PC = VI cos90° = 0 W.
Final Answer: Net average power absorbed by each pure L and pure C circuit over one complete cycle is 0 W.

Section 2: Resonance and Q-Factor

Resonance Derivation

At resonance, inductive reactance equals capacitive reactance.

XL = XCωL = 1/ωCω0 = 1/√LCf0 = 1/(2π√LC)

For a series LCR circuit, sharpness of resonance is:

Q = ω0L/R = (1/R)√(L/C)
f₀If

Question 6

Obtain the resonant frequency of a series LCR circuit with L = 2.0 H, C = 32 μF and R = 10 Ω. Also find the Q-factor.

JEE Main LevelMedium

Given: L = 2.0 H, C = 32 × 10-6 F, R = 10 Ω.

Formula: ω0 = 1/√LC f0 = ω0/2π Q = ω0L/R

  1. LC = 2 × 32 × 10-6 = 64 × 10-6.
  2. √LC = 8 × 10-3.
  3. ω0 = 1/(8 × 10-3) = 125 rad s-1.
  4. f0 = 125/(2π) = 19.9 Hz.
  5. Q = 125 × 2/10 = 25.
Final Answer: f0 = 19.9 Hz, Q = 25.

Question 9

A series LCR circuit with R = 20 Ω, L = 1.5 H and C = 35 μF is connected to a variable-frequency 200 V AC supply. At natural frequency, find average power in one cycle.

CBSE BoardMedium

At resonance: Z = R and cosφ = 1.

Formula: P = V²/R

  1. P = 200²/20.
  2. P = 40000/20 = 2000 W.
Final Answer: Average power = 2000 W.

Question 10

A radio tunes over 800 kHz to 1200 kHz. If L = 200 μH, find the required range of variable capacitance.

JEE Main LevelMedium

Formula: f = 1/(2π√LC), so C = 1/(4π²f²L)

  1. For f = 800 kHz, C = 1/[4π²(8 × 105)²(200 × 10-6)] = 1.98 × 10-10 F = 198 pF.
  2. For f = 1200 kHz, C = 8.80 × 10-11 F = 88 pF.
  3. Higher frequency needs lower capacitance.
Final Answer: C ranges from about 88 pF to 198 pF.

Question 11

A series LCR circuit is connected to a 230 V variable-frequency source with L = 5.0 H, C = 80 μF and R = 40 Ω. Determine resonant frequency, impedance/current amplitude at resonance and RMS voltage across R, L and C.

JEE Main LevelDifficult

Given: L = 5 H, C = 80 μF, R = 40 Ω, Vrms = 230 V.

  1. ω0 = 1/√(5 × 80 × 10-6) = 50 rad s-1.
  2. f0 = 50/(2π) = 7.96 Hz.
  3. At resonance, Z = R = 40 Ω.
  4. Irms = 230/40 = 5.75 A, so current amplitude I0 = √2 × 5.75 = 8.13 A.
  5. XL = ωL = 50 × 5 = 250 Ω and XC = 250 Ω.
  6. VR = IR = 230 V.
  7. VL = IXL = 5.75 × 250 = 1437.5 V.
  8. VC = 1437.5 V.
Final Answer: f0 = 7.96 Hz, Z = 40 Ω, I0 = 8.13 A, VR = 230 V, VL = VC = 1437.5 V rms.

Question 20

A series LCR circuit with L = 0.12 H, C = 480 nF and R = 23 Ω is connected to a 230 V variable-frequency supply. Determine maximum-current frequency, maximum current amplitude, maximum-power frequency, maximum power, half-power frequencies and Q-factor.

JEE Advanced LevelDifficult

Given: L = 0.12 H, C = 480 × 10-9 F, R = 23 Ω, Vrms = 230 V.

  1. ω0 = 1/√LC = 4166.7 rad s-1.
  2. f0 = 4166.7/(2π) = 663.1 Hz. This is frequency for maximum current and maximum average power.
  3. At resonance, Irms,max = V/R = 230/23 = 10 A.
  4. Current amplitude I0,max = √2 × 10 = 14.14 A.
  5. Pmax = V²/R = 230²/23 = 2300 W.
  6. At half power, |XL - XC| = R.
  7. Solving Lω² - Rω - 1/C = 0 gives f2 = 678.6 Hz.
  8. Solving Lω² + Rω - 1/C = 0 gives f1 = 648.1 Hz.
  9. Q = ω0L/R = 4166.7 × 0.12/23 = 21.7.
Final Answer: f0 = 663 Hz, I0,max = 14.14 A, Pmax = 2300 W, half-power frequencies = 648 Hz and 679 Hz, Q = 21.7.

Question 21

Find resonant frequency and Q-factor for L = 3.0 H, C = 27 μF, R = 7.4 Ω. Suggest a method to double sharpness of resonance.

JEE Main LevelDifficult
  1. ω0 = 1/√(3 × 27 × 10-6) = 111.1 rad s-1.
  2. f0 = 111.1/(2π) = 17.68 Hz.
  3. Q = ω0L/R = 111.1 × 3/7.4 = 45.0.
  4. Since Q ∝ 1/R for fixed L and C, to double Q, halve resistance.
  5. Required R = 7.4/2 = 3.7 Ω.
Final Answer: f0 = 17.7 Hz, Q = 45.0. To double sharpness, reduce R to 3.7 Ω.

Section 3: LC Oscillations

ω = 1/√LCT = 2π√LCUE = ½CV² = q²/(2C)UB = ½LI²

In an ideal LC circuit, energy oscillates between the capacitor's electric field and the inductor's magnetic field. Total energy remains constant.

Fully electric energy
Fully magnetic energy
Equal sharing

Question 7

A charged 30 μF capacitor is connected to a 27 mH inductor. Find angular frequency of free oscillations.

NEET LevelMedium

Given: C = 30 μF, L = 27 mH.

  1. LC = 27 × 10-3 × 30 × 10-6 = 8.1 × 10-7.
  2. √LC = 9 × 10-4.
  3. ω = 1/√LC = 1/(9 × 10-4) = 1111 rad s-1.
Final Answer: ω = 1.11 × 103 rad s-1.

Question 8

If initial charge on capacitor in Exercise 7 is 6 mC, find total energy stored initially and at a later time.

CBSE BoardMedium

Given: q = 6 mC = 6 × 10-3 C, C = 30 μF.

Formula: U = q²/(2C)

  1. U = (6 × 10-3)²/[2 × 30 × 10-6].
  2. U = 36 × 10-6 / 60 × 10-6 = 0.6 J.
  3. In an ideal LC circuit, total energy remains constant at all later times.
Final Answer: Initial energy = 0.6 J; later total energy = 0.6 J.

Question 12

An LC circuit contains L = 20 mH, C = 50 μF and initial charge 10 mC. Resistance is negligible. Determine energy, natural frequency, times for electric/magnetic/equal sharing and energy dissipated if resistor is introduced.

JEE Advanced LevelDifficult
  1. Initial energy U = q²/(2C) = (10 × 10-3)²/[2 × 50 × 10-6] = 1.0 J.
  2. ω = 1/√(20 × 10-3 × 50 × 10-6) = 1000 rad s-1.
  3. f = ω/2π = 159.15 Hz.
  4. T = 2π/ω = 6.283 × 10-3 s.
  5. Energy completely electric at t = 0, T/2, T ...
  6. Energy completely magnetic first at t = T/4 = 1.571 ms.
  7. Energy equally shared first at t = T/8 = 0.785 ms, then every T/4.
  8. If a resistor is introduced, energy is dissipated as heat; eventually the full initial 1.0 J is lost if oscillations die out completely.
Final Answer: U = 1.0 J, ω = 1000 rad s-1, f = 159 Hz, T = 6.28 ms, first magnetic maximum = 1.57 ms, first equal sharing = 0.785 ms.

Section 4: RL Circuits

XL = ωLZ = √(R² + XL²)I = V/Ztanφ = XL/R

At very high frequency, XL = 2πfL becomes very large, so current becomes very small. Thus an inductor behaves nearly as an open circuit.

V_R, IV_LV

Question 13

A coil of inductance 0.50 H and resistance 100 Ω is connected to a 240 V, 50 Hz AC supply. Find maximum current and time lag between voltage maximum and current maximum.

JEE Main LevelDifficult
  1. XL = 2π × 50 × 0.50 = 157.08 Ω.
  2. Z = √(100² + 157.08²) = 186.21 Ω.
  3. Irms = 240/186.21 = 1.289 A.
  4. Maximum current I0 = √2 × 1.289 = 1.82 A.
  5. tanφ = 157.08/100, so φ = 57.52° = 1.0039 rad.
  6. Time lag = φ/ω = 1.0039/(2π × 50) = 3.20 × 10-3 s.
Final Answer: I0 = 1.82 A; current maximum lags voltage maximum by 3.20 ms.

Question 14

Repeat Exercise 13 for a 240 V, 10 kHz supply. Explain why an inductor behaves almost like an open circuit at very high frequencies.

JEE Main LevelDifficult
  1. XL = 2π × 10000 × 0.50 = 31415.9 Ω.
  2. Z = √(100² + 31415.9²) = 31416.1 Ω.
  3. Irms = 240/31416.1 = 0.00764 A.
  4. I0 = √2 × 0.00764 = 0.0108 A.
  5. φ ≈ tan-1(31415.9/100) = 89.82°.
  6. Time lag = φ/ω = 2.49 × 10-5 s.
  7. At high f, XL becomes very large, so current becomes almost zero.
Final Answer: I0 = 0.0108 A, lag = 24.9 μs. Inductor behaves nearly as an open circuit because XL is very large.

Section 5: RC Circuits

XC = 1/ωCZ = √(R² + XC²)tanφ = XC/R

At very high frequency, XC becomes very small, so the capacitor offers very little opposition and behaves almost like a conductor. For DC, f = 0, so XC is infinite after charging.

V_R, IV_CV

Question 15

A 100 μF capacitor in series with a 40 Ω resistor is connected to a 110 V, 60 Hz supply. Determine maximum current and time lag between current maximum and voltage maximum.

JEE Main LevelDifficult
  1. XC = 1/(2π × 60 × 100 × 10-6) = 26.53 Ω.
  2. Z = √(40² + 26.53²) = 48.00 Ω.
  3. Irms = 110/48.00 = 2.292 A.
  4. I0 = √2 × 2.292 = 3.24 A.
  5. φ = tan-1(26.53/40) = 33.55° = 0.5856 rad.
  6. For RC, current leads voltage. Time lead = φ/ω = 0.5856/(2π × 60) = 1.55 ms.
Final Answer: I0 = 3.24 A; current maximum leads voltage maximum by 1.55 ms.

Question 16

Repeat Exercise 15 for a 110 V, 12 kHz supply. Explain why a capacitor behaves almost like a conductor at very high frequencies and compare with DC behaviour.

JEE Main LevelDifficult
  1. XC = 1/(2π × 12000 × 100 × 10-6) = 0.1326 Ω.
  2. Z = √(40² + 0.1326²) = 40.0002 Ω ≈ 40 Ω.
  3. Irms = 110/40.0002 = 2.75 A.
  4. I0 = √2 × 2.75 = 3.89 A.
  5. φ = tan-1(0.1326/40) = 0.190°.
  6. Time lead = φ/ω = 4.40 × 10-8 s.
  7. At high f, XC is almost zero. For DC, f = 0 and XC becomes infinite after charging.
Final Answer: I0 = 3.89 A, time lead = 4.40 × 10-8 s. Capacitor behaves almost like a conductor at high AC frequency but blocks steady DC.

Section 6: Parallel Resonance

For parallel R, L and C at resonance, branch currents in L and C are equal and opposite in phase. They cancel in the supply line, so total current is minimum and is mainly the resistor branch current.

XL = XCI = √[IR² + (IC - IL)²]
RLC

Question 17

Keeping the source frequency equal to the resonant frequency of the series LCR circuit, arrange L, C and R in parallel and show that total current is minimum. Find RMS current in each branch.

JEE Advanced LevelDifficult

Using the values from Question 11: V = 230 V, R = 40 Ω, L = 5 H, C = 80 μF and ω0 = 50 rad s-1.

  1. XL = ωL = 50 × 5 = 250 Ω.
  2. XC = 1/(ωC) = 1/(50 × 80 × 10-6) = 250 Ω.
  3. IR = V/R = 230/40 = 5.75 A.
  4. IL = V/XL = 230/250 = 0.92 A lagging.
  5. IC = V/XC = 230/250 = 0.92 A leading.
  6. IL and IC cancel because they are equal and opposite, so supply current is minimum and equals IR.
Final Answer: IR = 5.75 A, IL = 0.92 A, IC = 0.92 A; net current is minimum = 5.75 A.

Section 7: Power in AC Circuits

P = VIcosφP = I²RP = V²/R

Active power P is useful power in watts. Reactive power Q oscillates between source and reactive elements. Apparent power S = VI is measured in VA.

PQSφ

Question 18

A circuit containing an 80 mH inductor and a 60 μF capacitor in series is connected to a 230 V, 50 Hz supply. Resistance is negligible. Find currents, voltages and average powers.

JEE Main LevelDifficult
  1. ω = 2π × 50 = 314.16 rad s-1.
  2. XL = 314.16 × 0.080 = 25.13 Ω.
  3. XC = 1/(314.16 × 60 × 10-6) = 53.05 Ω.
  4. Net reactance = XL - XC = -27.92 Ω, so |Z| = 27.92 Ω.
  5. Irms = 230/27.92 = 8.24 A; I0 = 11.65 A.
  6. VL = IXL = 8.24 × 25.13 = 207 V.
  7. VC = IXC = 8.24 × 53.05 = 437 V.
  8. Average power in ideal L = 0, in ideal C = 0, total = 0 because R = 0.
Final Answer: I0 = 11.65 A, Irms = 8.24 A, VL = 207 V, VC = 437 V, all average powers = 0 W.

Question 19

Repeat Exercise 18 if circuit has resistance 15 Ω. Find average power absorbed by each element and total power.

JEE Main LevelDifficult

Use XL = 25.13 Ω, XC = 53.05 Ω, net X = -27.92 Ω.

  1. Z = √(15² + 27.92²) = 31.69 Ω.
  2. Irms = 230/31.69 = 7.26 A; I0 = 10.26 A.
  3. Power in resistor = I²R = 7.26² × 15 = 790 W.
  4. Average power in ideal inductor = 0 W.
  5. Average power in ideal capacitor = 0 W.
  6. Total average power = 790 W.
Final Answer: PR = 790 W, PL = 0 W, PC = 0 W, total power = 790 W.

Section 8: Transformer Applications

Vs/Vp = Ns/NpIs/Ip = Np/NsPloss = I²R

Step-up transformer increases voltage and decreases current. Step-down transformer decreases voltage and increases current. High-voltage transmission reduces current, so I²R loss becomes small.

Step-up
Step-down
High V, low I, low I²R loss

Question 23

A power transmission line feeds a step-down transformer with primary winding of 4000 turns. Output voltage is 230 V when input voltage is 2300 V. Find secondary turns.

CBSE BoardMedium

Formula: Vs/Vp = Ns/Np

  1. Ns = Np × Vs/Vp.
  2. Ns = 4000 × 230/2300 = 400.
Final Answer: Secondary turns = 400.

Question 24

In a hydroelectric plant, water head is 300 m and flow rate is 100 m³/s. Generator efficiency is 60%. Estimate electric power generated.

JEE Main LevelMedium

Formula: P = ηρghQ

  1. ρ = 1000 kg m-3, g = 9.8 m s-2, h = 300 m, Q = 100 m³ s-1, η = 0.60.
  2. P = 0.60 × 1000 × 9.8 × 300 × 100.
  3. P = 1.764 × 108 W.
Final Answer: Electric power generated = 176 MW approximately.

Question 25

A town requiring 800 kW at 220 V is 15 km from a station generating at 440 V. Transmission line resistance is 0.5 Ω/km and power is delivered through a 4000-220 V transformer. Determine line loss, power supplied and step-up transformer characteristics.

JEE Advanced LevelDifficult

Assumption: Transmission uses two conductors, so total wire length = 30 km. Total line resistance = 0.5 × 30 = 15 Ω.

  1. Load power = 800 kW. Transmission voltage = 4000 V.
  2. Line current I = P/V = 800000/4000 = 200 A.
  3. Line loss = I²R = 200² × 15 = 600000 W = 600 kW.
  4. Power supplied by plant = load power + line loss = 800 kW + 600 kW = 1400 kW.
  5. Plant voltage = 440 V and transmission voltage = 4000 V, so the plant transformer is step-up.
  6. Voltage ratio = 4000/440 = 9.09, so Ns/Np = 9.09.
  7. Primary current of step-up transformer ≈ 1.4 × 106/440 = 3182 A.
  8. Secondary current = line current = 200 A.
Final Answer: Line loss = 600 kW, plant supplies 1400 kW, step-up transformer ratio = 440 V to 4000 V, turns ratio = 1 : 9.09.

If a textbook interprets 0.5 Ω/km as already the loop resistance per km, then R = 7.5 Ω and loss = 300 kW. The two-wire interpretation is used here because the town is 15 km away.

Question 26

Repeat Exercise 25 using a 40,000-220 V transformer and explain why high-voltage transmission is preferred.

JEE Advanced LevelDifficult
  1. Transmission voltage = 40000 V.
  2. Line current I = 800000/40000 = 20 A.
  3. Using total line resistance 15 Ω, line loss = 20² × 15 = 6000 W = 6 kW.
  4. Power supplied by plant = 800 kW + 6 kW = 806 kW.
  5. Step-up transformer at plant: 440 V to 40000 V.
  6. Turns ratio Ns/Np = 40000/440 = 90.9.
  7. Compared with Question 25, current falls from 200 A to 20 A, so I²R loss falls by factor 100.
Final Answer: Line loss = 6 kW, plant supplies 806 kW, step-up ratio = 1 : 90.9. High voltage is preferred because it greatly reduces current and hence I²R loss.

Section 9: Conceptual Questions

Question 22(a-e)

Answer the conceptual questions.

CBSE BoardNEET/JEE Theory

(a) Is applied instantaneous voltage equal to algebraic sum of voltages across series elements?

Theory: Kirchhoff's loop law is valid instantaneously, but in AC phasor analysis, RMS voltage magnitudes across R, L and C cannot be added algebraically because they have phase differences.

Exam Answer: Instantaneous voltages add algebraically, but RMS phasor magnitudes must be added vectorially.

One-line: Add instantaneous values algebraically; add RMS phasors vectorially.

(b) Why is a capacitor used in primary circuit of an induction coil?

Theory: The capacitor reduces sparking at the break and makes current fall rapidly, producing a large induced emf in secondary.

Exam Answer: It prevents sparking and increases rate of change of current.

One-line: Capacitor gives sharp current break and high induced emf.

(c) Show that in LC series circuit with DC and AC components, DC appears across C while AC appears across L.

Theory: For DC, f = 0, so XL = 0 and XC = ∞. Hence DC is blocked by capacitor and appears across C. For high-frequency AC, XL is large and XC is small, so AC mainly appears across L.

Exam Answer: DC drops across C; AC drops across L due to frequency-dependent reactance.

One-line: C blocks DC, L opposes AC.

(d) What happens when a choke coil is inserted with a lamp on AC and DC lines?

Theory: On AC, choke has high inductive reactance and reduces current, so lamp becomes dim. On DC after steady state, only winding resistance remains, so effect is small.

Exam Answer: Lamp dims on AC; on DC it glows almost normally except for small resistance drop.

One-line: Choke opposes AC strongly but DC weakly.

(e) Why is choke coil preferred over ordinary resistor in fluorescent lamp circuits?

Theory: A choke limits current mainly by inductive reactance, so it consumes very little average power compared with a resistor, which wastes energy as heat.

Exam Answer: Choke limits current with small power loss.

One-line: Choke saves power; resistor wastes heat.

Section 10: Formula Sheet

AC Formulas

Vrms=V₀/√2Irms=I₀/√2Vavg=2V₀/π

RL Formulas

XL=ωLZ=√(R²+XL²)tanφ=XL/R

RC Formulas

XC=1/ωCZ=√(R²+XC²)tanφ=XC/R

LC Formulas

ω=1/√LCT=2π√LCU=q²/(2C)

LCR and Resonance

Z=√[R²+(XL-XC)²]f₀=1/(2π√LC)Q=ω₀L/R

Transformer

Vs/Vp=Ns/NpIs/Ip=Np/NsPloss=I²R

Section 11: NEET / JEE / CBSE PYQs Classification

LevelQuestionsReason
CBSE Board Level1, 2, 5, 8, 9, 22, 23Direct formulas and conceptual board-style answers.
NEET Level2, 3, 4, 7, 13, 15Formula-based AC current and phase questions.
JEE Main Level6, 10, 11, 14, 16, 18, 19, 21, 24Multi-step calculations using reactance, resonance and power.
JEE Advanced Level12, 17, 20, 25, 26Concept linking, half-power frequencies, LC timing and transmission efficiency.

Section 12: Common Mistakes

RMS vs Peak

Use √2 correctly. Power formulas use RMS values unless peak values are explicitly required.

Resonance

At resonance, XL = XC, not L = C. Also Z = R only for series LCR.

Transformer

Step-up voltage means current decreases. Do not assume both increase.

Power Factor

Pure L and C have zero average power because cos90° = 0.

Inductive Reactance

XL increases with frequency. Inductors block high-frequency AC strongly.

Capacitive Reactance

XC decreases with frequency. Capacitors pass high-frequency AC more easily but block steady DC.

Section 13: Exam Tips by Kumar Sir

Need Help in Alternating Current and LCR Circuits?

If AC circuits, LC oscillations, resonance, transformers, power factor, phasor diagrams or numerical problems are not crystal clear, learn Physics through one-to-one personalized guidance from Kumar Sir.

Contact: +91-9958461445

Website: KumarPhysicsClasses.com

Email: kumarsirphysics@gmail.com

Scroll to Top