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JEE Mains Physics · Formula sheet

Alternating Current formulas

12 formulas, 1 reference table and 35 common traps for JEE Mains Physics Alternating Current, grouped by subtopic.

Full notes with worked examples

RMS Values and Timing in AC

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RMS value, meters and timing on a sine wave

RMS value and time on the wave

Irms=I02t=Δ(ωt)ωI_{rms} = \frac{I_0}{\sqrt{2}} \qquad t = \frac{\Delta(\omega t)}{\omega}

RMS of a sum: d.c. plus a.c., sine plus cosine

Mean squares add

Irms=Idc2+I022Irms2=1T∫0Ti2 dtI_{rms} = \sqrt{I_{dc}^{2} + \frac{I_0^{2}}{2}} \qquad I_{rms}^{2} = \frac{1}{T}\int_0^{T} i^{2}\,dt

Common traps

Peak to rms is T/8, not T/4

From the peak, the wave must lose a phase of π/4 to fall to 1/√2 of the peak. That is one eighth of a cycle. A quarter cycle takes it all the way to zero.

Taking the coefficient of t as the frequency

In sin(200πt) the coefficient 200π is ω in rad/s. The frequency is ω/2π = 100 Hz. Do not multiply by 2π again.

Using the peak where the meter reads rms

A meter reading or a rating like '220 V' is an rms value. Multiply by √2 only when the question asks for a peak.

Adding rms values

For 3 + 4√2 sin ωt the rms is √(9 + 16) = 5, not 3 + 4 = 7. Mean squares add; rms values do not.

Averaging i instead of i²

The plain average of a sinusoid over a cycle is zero. The rms comes from the average of the SQUARE, and the root is taken last.

Adding the amplitudes of a sine and a cosine

sin ωt and cos ωt are 90° apart, so their amplitudes combine like the sides of a right triangle: √(I₁² + I₂²), not I₁ + I₂.

Reactance of a Resistor, Inductor and Capacitor

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Reactance: ωL and 1/ωC

Reactances

XL=ωL=2πfLXC=1ωC=12πfCX_L = \omega L = 2\pi fL \qquad X_C = \frac{1}{\omega C} = \frac{1}{2\pi fC}

A real coil on d.c. and a.c., and the frequency limits

Coil on d.c. and on a.c.

R=VdcIdcZ=VrmsIrms=R2+XL2R = \frac{V_{dc}}{I_{dc}} \qquad Z = \frac{V_{rms}}{I_{rms}} = \sqrt{R^{2} + X_L^{2}}

Phase of the current in R, L and C

ElementOppositionChange with frequencyCurrent compared with voltageAverage power
Pure resistorRNoneIn phaseVrmsIrmsV_{rms}I_{rms}
Pure inductorXL=ωLX_L = \omega LGrows in proportion to fLags by π2\dfrac{\pi}{2}Zero
Pure capacitorXC=1ωCX_C = \dfrac{1}{\omega C}Falls as 1f\dfrac{1}{f}Leads by π2\dfrac{\pi}{2}Zero
Ideal L and C in series∣XL−XC∣|X_L - X_C|Falls to zero at resonanceLags by π2\dfrac{\pi}{2} if XL>XCX_L > X_C, leads if XC>XLX_C > X_LZero
No resistance anywhere, so the gap is exactly 90° whichever reactance wins.
Series LCRR2+(XL−XC)2\sqrt{R^{2} + (X_L - X_C)^{2}}Least at resonanceAngle ϕ\phi with tan⁡ϕ=XL−XCR\tan\phi = \dfrac{X_L - X_C}{R}VrmsIrmscos⁡ϕV_{rms}I_{rms}\cos\phi
Current zero while the voltage is at its peak means a 90° gap: no resistance in the circuit.

Common traps

Making the capacitive reactance rise with frequency

Capacitive reactance goes DOWN as f or C goes up. Doubling both divides it by 4.

Multiplying a given ω by 2π

In sin(500t) the 500 is already ω in rad/s. Use 2πf only when the frequency is given in hertz.

Peak or rms

An emf written as E₀ sin ωt gives the PEAK current. An a.c. ammeter shows the rms value, √2 times smaller.

Which one leads in an inductor

In an inductor the VOLTAGE leads the current. Saying 'the current leads' is the capacitor's rule. CIVIL settles it.

Adding 90° to the wrong quantity

Going from current to voltage in an inductor, add π/2. Going from voltage to current, subtract it. For a capacitor it is the other way round.

Taking V/I on a.c. as the reactance

V/I on a.c. is the impedance, which still contains the coil's resistance. Find R on d.c. first, then XL=Z2−R2X_L = \sqrt{Z^{2} - R^{2}}.

Power in the inductance

Only the resistance takes power: P = I²R with the rms current. I²Z overstates it.

Average or peak stored energy

½LI² with the rms current is the energy averaged over a cycle; with the peak current it is the maximum. The two differ by a factor of 2.

Series LCR: Impedance, Phase and Power

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Impedance and the voltage triangle

Impedance and voltages

Z=R2+(XL−XC)2V2=VR2+(VL−VC)2Z = \sqrt{R^{2} + (X_L - X_C)^{2}} \qquad V^{2} = V_R^{2} + (V_L - V_C)^{2}

Phase angle and power factor

tan⁡ϕ=XL−XCRcos⁡ϕ=RZ\tan\phi = \frac{X_L - X_C}{R} \qquad \cos\phi = \frac{R}{Z}

Average power, wattless current and the choke coil

Average power

P=VrmsIrmscos⁡ϕ=12V0I0cos⁡ϕ=Irms2RP = V_{rms}I_{rms}\cos\phi = \frac{1}{2}V_0I_0\cos\phi = I_{rms}^{2}R

Common traps

Adding the voltages as numbers

VR+VL+VCV_R + V_L + V_C is not the supply voltage. The parts are out of phase, so use V2=VR2+(VL−VC)2V^{2} = V_R^{2} + (V_L - V_C)^{2}.

Changing only one reactance with the frequency

When ω changes, XLX_L and XCX_C both change, in opposite directions. Scale both before finding Z.

Mixing peak and rms

Peak voltage over Z gives the peak current; rms over Z gives rms. A 'percent lower' frequency means ω times the remaining fraction: 30% lower is 0.7ω.

Using sin φ for the power factor

The power factor is cos φ = R/Z. The ratio of the net reactance to Z is sin φ.

Losing lead or lag

cos φ is the same for +φ and −φ. Decide lead or lag from which reactance is bigger, or from which equation has the larger phase constant.

Letting the amplitude change the power factor

A new source amplitude changes the current, not the angle. Only a new frequency or a new part changes the power factor.

Peak values in the rms formula

VrmsIrmscos⁡ϕV_{rms}I_{rms}\cos\phi needs rms values. With peak values, halve the product: 12V0I0cos⁡ϕ\dfrac{1}{2}V_0I_0\cos\phi.

Using Z in I²R

Power is Irms2RI_{rms}^{2}R. Writing I2ZI^{2}Z charges the inductor and capacitor for power they never keep.

Resonance, Quality Factor and Bandwidth

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The resonant frequency

Resonant frequency

ω0=1LCf0=12πLC\omega_0 = \frac{1}{\sqrt{LC}} \qquad f_0 = \frac{1}{2\pi\sqrt{LC}}

Current, impedance and power at resonance

At resonance

Z=RI0=V0RPmax=Vrms2RZ = R \qquad I_0 = \frac{V_0}{R} \qquad P_{max} = \frac{V_{rms}^{2}}{R}

Quality factor and bandwidth

Q=ω0LR=1RLCΔω=RL=ω0QQ = \frac{\omega_0L}{R} = \frac{1}{R}\sqrt{\frac{L}{C}} \qquad \Delta\omega = \frac{R}{L} = \frac{\omega_0}{Q}

Common traps

ω or f

1LC\dfrac{1}{\sqrt{LC}} is ω in rad/s. For hertz, divide by 2π. Check which one the blank asks for.

Putting R into the formula

The resistance changes how tall and wide the resonance peak is, never where it sits.

Losing a power of ten

Convert first: 1 μF=10−6 F1\ \mu\text{F} = 10^{-6}\ \text{F}, 1 nF=10−9 F1\ \text{nF} = 10^{-9}\ \text{F}, 1 mH=10−3 H1\ \text{mH} = 10^{-3}\ \text{H}. Most wrong answers here are off by a power of ten.

Forgetting √2 on the amplitude

A supply quoted as '220 V' is rms. At resonance V/R is then the rms current; the amplitude is √2 times that.

Using L and C to find the resonant current

Once the circuit is at resonance, L and C have cancelled. The current depends only on V and R.

Reading the I–ω curve the wrong way round

On the low-frequency side XCX_C is large, so the circuit is capacitive. On the high side XLX_L wins.

Half power is not half current

At the edges of the band the CURRENT is 1/√2 of its peak. Power goes as I², so the power is half.

Q with f₀ instead of ω₀

Q = ω₀L/R uses the angular frequency. Using f₀ makes Q too small by a factor of 2π.

Bandwidth in rad/s or in Hz

R/L is a bandwidth in rad/s. In hertz it is R/(2πL).

LC Oscillations, Transformers and AC Devices

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LC oscillations

LC oscillator

ω=1LCImax=Q0LC=V0CL\omega = \frac{1}{\sqrt{LC}} \qquad I_{max} = \frac{Q_0}{\sqrt{LC}} = V_0\sqrt{\frac{C}{L}}

Transformers and a.c. devices

Transformer

VsVp=NsNpVsIs=η VpIp\frac{V_s}{V_p} = \frac{N_s}{N_p} \qquad V_sI_s = \eta\,V_pI_p

Common traps

√(C/L) or √(L/C)

From ½CV₀² = ½LI², the current is V₀√(C/L). A large capacitor or a small inductor gives a large current.

Charge fraction and energy fraction

Energy goes as q². Half the charge leaves a quarter of the energy in the capacitor, not half.

Turning the ratio upside down

Voltage follows the turns: more turns, more volts. Current goes the other way. Write VsVp=NsNp\dfrac{V_s}{V_p} = \dfrac{N_s}{N_p} before putting numbers in.

Forgetting the efficiency on the current

With losses, the output power is η times the input. Find the output current from that power, not from the ideal turns ratio.

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