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Alternating Currents Formulas for JEE
Alternating Currents formulas for JEE help you solve questions on RMS values, reactance, impedance, LCR circuits, resonance, AC power and transformers. This formula sheet brings together key relations, clear tables, worked examples and useful shortcuts for JEE Main and Advanced preparation. Each section explains when to apply a formula and highlights common mistakes to avoid.
Use this sheet alongside JEE Alternating Currents PYQs and Physics mock tests to strengthen your understanding and improve accuracy. Pay attention to phase differences, units and peak versus RMS values while practising. Before the exam, revisit the quick revision section to refresh the important formulas and concepts.
Notation
| Symbol | Stands for | Units |
|---|---|---|
| $$i$$ | Instantaneous current | A |
| $$I_0$$ | Peak (amplitude) of current | A |
| $$I_{\mathrm{rms}}$$ | Root-mean-square current | A |
| $$e$$ or $$v$$ | Instantaneous emf / voltage | V |
| $$E_0$$ | Peak emf | V |
| $$E_{\mathrm{rms}}$$ | RMS emf | V |
| $$\omega$$ | Angular frequency $$2\pi f$$ | rad s-1 |
| $$f$$ | Frequency | Hz |
| $$T$$ | Time period | s |
| $$R$$ | Resistance | Ω |
| $$L$$ | Inductance | H |
| $$C$$ | Capacitance | F |
| $$X_L$$ | Inductive reactance | Ω |
| $$X_C$$ | Capacitive reactance | Ω |
| $$Z$$ | Impedance | Ω |
| $$\phi$$ | Phase angle between $$i$$ and $$v$$ | rad |
| $$P$$ | Average power | W |
| $$Q$$ | Quality factor | dimensionless |
| $$\Delta\omega$$ | Bandwidth | rad s-1 |
| $$V_1, V_2$$ | Primary and secondary rms voltages of a transformer | V |
| $$N_1, N_2$$ | Number of turns in primary and secondary | — |
| $$I_1, I_2$$ | Primary and secondary rms currents | A |
Phasors, Angular Frequency and RMS Relations Formulas
This sub-topic gives you the language of AC: how to write $$i$$ and $$v$$ as rotating vectors (phasors) and jump between peak and RMS. Most one-step “find the RMS value” MCQs live here and the phasor picture feeds directly into impedance in the next sub-topic.
Formulas
| Formula | What it gives you | Valid when |
|---|---|---|
| $$i = I_0 \sin(\omega t + \phi_i)$$ | Instantaneous current | Any single-frequency AC |
| $$v = E_0 \sin(\omega t + \phi_v)$$ | Instantaneous emf | — |
| $$I_{\mathrm{rms}} = I_0/\sqrt2$$ | RMS current | Pure sine wave |
| $$E_{\mathrm{rms}} = E_0/\sqrt2$$ | RMS voltage | Pure sine wave |
| $$\omega = 2\pi f = 2\pi/T$$ | Angular frequency | — |
| $$\text{Phasor length} = \text{RMS value}$$ | Scale choice in phasor diagrams | JEE convention |
Worked Example
An AC source is given by $$v=170\sin(314 t)$$ volts. Find (a) the frequency, (b) the RMS value of voltage.
- Compare with $$v = E_0 \sin(\omega t)$$ to get $$E_0 = 170\ \mathrm{V}, \quad \omega = 314\ \mathrm{rad\,s^{-1}}$$.
- (a) $$f=\omega/2\pi=314/2\pi \approx 50\ \text{Hz}$$.
- (b) $$E_{\mathrm{rms}}=E_0/\sqrt2 = 170/1.414 \approx 120.2\ \text{V}$$.
Answer: (a) 50 Hz, (b) 120 V
Shortcuts and Special Cases
- Always memorise: $$\sqrt2 \approx 1.414$$; $$1/\sqrt2 \approx 0.707$$.
- If the examiner gives peak-to-peak value $$E_{\text{pp}}$$, remember $$E_0 = E_{\text{pp}}/2$$.
Watch Out
- Writing $$I_{\mathrm{rms}} = I_0 \sqrt2$$—reverse of the correct relation.
- Quoting phasor length as peak value; in JEE the phasor radius equals RMS.
- Mixing degrees and radians when using $$\omega t$$ inside the sine.
Pure R, Pure L, Pure C under AC Formulas
Here you practise the simplest phase relations: current in phase with voltage for a resistor, lags by $$90^\circ$$ in an inductor, leads by $$90^\circ$$ in a capacitor. These relations become the building blocks for mixed circuits next.
Formulas
| Formula | What it gives you | Valid when |
|---|---|---|
| Resistor: $$v=Ri$$ | Ohm’s law (instantaneous and RMS) | Pure $$R$$ only |
| Inductor: $$X_L = \omega L$$ | Inductive reactance | Ideal $$L$$, no $$R$$ |
| Inductor: $$v = L\,di/dt$$ | Instantaneous relation | — |
| Capacitor: $$X_C = 1/(\omega C)$$ | Capacitive reactance | Ideal $$C$$ |
| Capacitor: $$i = C\,dv/dt$$ | Instantaneous relation | — |
| Phase (R): $$\phi=0$$ | Current and voltage in phase | Pure $$R$$ |
| Phase (L): $$\phi=+90^\circ$$ | Current lags voltage | Pure $$L$$ |
| Phase (C): $$\phi=-90^\circ$$ | Current leads voltage | Pure $$C$$ |
Worked Example
A 10-mH ideal inductor is connected to a 200-V (rms), 50-Hz supply. Calculate the rms current.
- Compute $$X_L = \omega L = 2\pi f L = 2\pi(50)(0.01)= 3.14\ \Omega$$.
- RMS current $$I_{\mathrm{rms}} = E_{\mathrm{rms}}/X_L = 200/3.14 \approx 63.7\ \text{A}$$.
Answer: 64 A (to two significant figures)
Shortcuts and Special Cases
- At $$f=50$$ Hz, $$X_L (\Omega) \approx 314 L(\text{H})$$—handy mental multiply.
- At the same frequency, $$X_C (\Omega) \approx 1/(314 C(\text{F}))$$.
Watch Out
- Forgetting that reactance has the same unit as resistance.
- Using $$X_L = L/\omega$$ instead of $$\omega L$$.
- Assuming power is non-zero for pure L or C; it is zero because $$\cos\phi=0$$.
Series R-L, R-C and L-C Circuits Formulas
Now you mix two elements. The impedance triangle appears and so does a non-trivial phase angle. Most JEE Alternating Currents Questions that ask “find current and phase difference” come from here when they’re not about resonance.
Formulas
| Formula | What it gives you | Valid when |
|---|---|---|
| R-L: $$Z = \sqrt{R^2 + X_L^2}$$ | Impedance magnitude | Series $$R$$ and $$L$$ |
| R-L phase: $$\tan\phi = X_L/R$$ | Current lags by $$\phi$$ | — |
| R-C: $$Z = \sqrt{R^2 + X_C^2}$$ | Impedance | Series $$R$$ and $$C$$ |
| R-C phase: $$\tan\phi = -X_C/R$$ | Current leads by $$|\phi|$$ | — |
| L-C (no R): $$Z = |X_L - X_C|$$ | Net reactance | Ideal L and C in series |
Worked Example
A series circuit has $$R=40\ \Omega$$ and $$L=0.2\ \text{H}$$. It is connected to 230 V (rms), 50 Hz. Find (a) impedance, (b) rms current, (c) phase angle.
- $$X_L = \omega L = 2\pi(50)(0.2)=62.8\ \Omega$$.
- Impedance $$Z = \sqrt{40^2 + 62.8^2} = \sqrt{1600 + 3945} \approx \sqrt{5545} \approx 74.5\ \Omega$$.
- RMS current $$I = 230/74.5 \approx 3.09\ \text{A}$$.
- Phase $$\tan\phi = X_L/R = 62.8/40 = 1.57 \Rightarrow \phi \approx 57^\circ$$ (lag).
Answer: (a) 74.5 Ω, (b) 3.1 A, (c) 57° lag
Shortcuts and Special Cases
- If $$X_L \gg R$$, then $$Z \approx X_L$$ and $$\phi \approx 90^\circ$$—often enough to eliminate options.
- For R-C with $$X_C=R$$, current leads voltage by exactly $$45^\circ$$.
Watch Out
- Mixing sign convention for $$X_C$$; remember $$X_C$$ is positive by magnitude, the sign is absorbed in phase.
- Taking arithmetic instead of vector sum: $$Z \neq R + X_L$$.
- Using degree mode on calculator for $$\tan^{-1}$$ when you need radians for substitution elsewhere.
Series LCR Circuit and Resonance Formulas
This is the heart of AC for JEE. Expect a four-mark integer or MCQ about resonance frequency, quality factor, or voltage magnification. It links directly to bandwidth and filters in the next sub-topic.
Formulas
| Formula | What it gives you | Valid when |
|---|---|---|
| Resonance condition: $$X_L = X_C \Rightarrow \omega_0 = 1/\sqrt{LC}$$ | Angular resonance frequency | Series LCR |
| $$Z_{\min} = R$$ | Impedance at resonance | — |
| $$I_{\max} = E_{\mathrm{rms}}/R$$ | Current at resonance | — |
| Voltage across L or C: $$V_L = I_{\max} X_L, \quad V_C = I_{\max} X_C$$ | Voltage magnification | At resonance |
| Quality factor: $$Q = \omega_0 L / R = 1/(R\omega_0 C)$$ | Sharpness of resonance | Series circuit |
| Bandwidth: $$\Delta\omega = \omega_0/Q$$ | Range between half-power points | High-Q approximation |
| Half-power frequencies: $$\omega_{\pm} = \omega_0 \pm \tfrac12 \Delta\omega$$ | Frequencies where power halves | — |
Worked Example
An LCR circuit has $$L = 50\ \text{mH},\ C = 2\ \mu\text{F},\ R = 10\ \Omega$$. (a) Find the resonant frequency in Hz. (b) Calculate the quality factor.
- $$\omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{(50 \times 10^{-3})(2 \times 10^{-6})}} = \frac{1}{\sqrt{10^{-7}}} \approx 3162\ \mathrm{rad\,s^{-1}}$$.
- $$f_0 = \omega_0/2\pi = 3162 / 6.283 \approx 503\ \text{Hz}$$.
- Quality factor $$Q = \omega_0 L / R = 3162 (0.05)/10 = 15.8$$.
Answer: (a) 5.0 × 102 Hz, (b) 16
Shortcuts and Special Cases
- At resonance, power factor is unity—immediately sets $$P = E_{\mathrm{rms}} I_{\mathrm{rms}}$$.
- For high-Q circuits, $$\Delta f \approx f_0/Q$$ is easier to remember than the angular form.
- Voltage magnification equals $$Q$$; useful for quick checks.
Watch Out
- Plugging $$R$$ into the resonance formula—$$\omega_0$$ depends only on $$L$$ and $$C$$.
- Forgetting to convert $$\mu\text{F}$$ and $$\text{mH}$$ to SI before calculation.
- Using $$\sqrt{LC}$$ instead of $$1/\sqrt{LC}$$.
Parallel LCR Circuits Formulas
Parallel resonance is rarer but shows up in assertion-reason or subjective questions. The key difference: impedance peaks, current dips at resonance—opposite of series. Know the symmetric formulas so you do not misapply the series ones.
Formulas
| Formula | What it gives you | Valid when |
|---|---|---|
| Admittance: $$Y = \sqrt{G^2 + (B_L - B_C)^2}$$ | Magnitude, where $$G=1/R$$ | Parallel LCR |
| Parallel resonance: $$\omega_0 = 1/\sqrt{LC}$$ | Same numeric as series | Ideal components |
| Impedance at resonance: $$Z_{\max} = R$$ | Maximum impedance | — |
| Current minimum: $$I_{\min} = E_{\mathrm{rms}}/R$$ | Supply current | At resonance |
Worked Example
A parallel circuit has $$R=100\ \Omega, L=0.5\ \text{H}, C=20\ \mu\text{F}$$ connected to 200 V (rms). Find the supply current at resonance.
- The resonant angular frequency is the same as for a series LCR circuit: $$\omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{0.5 \times 20 \times 10^{-6}}} = \frac{1}{\sqrt{10^{-5}}} \approx 316\ \mathrm{rad\,s^{-1}}$$ (for verification only).
- At resonance, impedance $$Z_{\max}=R=100\ \Omega$$.
- Current $$I_{\min}=E_{\mathrm{rms}}/R=200/100=2\ \text{A}$$.
Answer: 2 A
Shortcuts and Special Cases
- Series: current max, impedance min. Parallel: reverse. Repeat until burnt in.
Watch Out
- Confusing conductance $$G$$ with admittance $$Y$$; $$Y$$ is not additive in magnitude.
- Quoting current max instead of min at parallel resonance.
Power in AC Circuits and Power Factor Formulas
Once you know $$I_{\mathrm{rms}}$$, $$E_{\mathrm{rms}}$$, and $$\phi$$, the examiner can ask for power and wattless current. Plenty of JEE Mains Formula based numericals live here because the arithmetic is quick but concept trips are many.
Formulas
| Formula | What it gives you | Valid when |
|---|---|---|
| Average power: $$P = E_{\mathrm{rms}} I_{\mathrm{rms}} \cos\phi$$ | True power delivered | Any single-frequency AC |
| Power factor: $$\cos\phi = R/Z$$ | Ratio of real to apparent power | Series circuit |
| Reactive power: $$Q_{\text{var}} = E_{\mathrm{rms}} I_{\mathrm{rms}} \sin\phi$$ | VArs (volt-ampere reactive) | — |
| Apparent power: $$S = E_{\mathrm{rms}} I_{\mathrm{rms}}$$ | VA rating | — |
| Wattless current: $$I_w = I_{\mathrm{rms}}\sin\phi$$ | Component that does no work | — |
Worked Example
A series R-C circuit draws 5 A from a 120 V (rms) source. The phase angle is $$37^\circ$$ (current leads). Find (a) average power, (b) wattless current.
- Average power $$P = E I \cos\phi = 120 \times 5 \times \cos(37^\circ) = 600 \times 0.8 = 480\ \text{W}$$.
- Wattless current $$I_w = I \sin\phi = 5 \times \sin(37^\circ) = 5 \times 0.6 = 3\ \text{A}$$.
Answer: (a) 480 W, (b) 3 A
Shortcuts and Special Cases
- Right-angle triangle for $$\cos 37^\circ = 4/5$$ and $$\sin 37^\circ = 3/5$$ saves calculator time.
- If $$P = 0$$, circuit is purely reactive: either $$X_L = X_C$$ with no $$R$$ or just L/C alone.
Watch Out
- Writing $$P=E I \sin\phi$$—that’s reactive power, not real.
- Mistaking lag for lead: sign does not affect $$\cos\phi$$ but shows up in phasor drawing.
Quality Factor, Bandwidth and Selectivity Formulas
This sub-topic refines the resonance discussion. Paper setters love asking you to compare two circuits on the basis of their $$Q$$ or bandwidth. It ties into communication filters, an interdisciplinary angle.
Formulas
| Formula | What it gives you | Valid when |
|---|---|---|
| $$Q = \omega_0 / \Delta\omega$$ | Defines quality factor | Any resonant circuit |
| $$\Delta f = f_2 - f_1$$ | Bandwidth between half-power frequencies | — |
| Selection ratio: $$f_0/\Delta f$$ | Measure of selectivity | — |
| Voltage magnification (series): $$V_L/E = Q$$ | Peak across L or C | At resonance |
Worked Example
A series resonant circuit has $$Q=50$$ and resonates at 1 MHz. Determine the bandwidth in kHz and the half-power frequencies.
- Bandwidth $$\Delta f = f_0/Q = 1\ \text{MHz} / 50 = 20\ \text{kHz}$$.
- Half-power: $$f_1 = f_0 - \Delta f/2 = 1.000\ \text{MHz} - 10\ \text{kHz} = 990\ \text{kHz}$$.
- Similarly, $$f_2 = 1.010\ \text{MHz}$$.
Answer: Bandwidth 20 kHz, $$f_1=990$$ kHz, $$f_2=1.010$$ MHz
Shortcuts and Special Cases
- High-$$Q$$ ⇒ narrow bandwidth; low-$$Q$$ ⇒ broad—state this if the question is conceptual.
Watch Out
- Confusing kHz with k rad/s in answers; always specify unit.
- Using $$Q = \Delta\omega/\omega_0$$—inverse of correct relation.
Transformers and AC Instruments Formulas
Exam setters slot quick numerical checks on turn ratio or efficiency under AC. Also, knowing how AC ammeters/voltmeters read RMS clarifies theory MCQs.
Formulas
| Formula | What it gives you | Valid when |
|---|---|---|
| Turn ratio: $$E_2/E_1 = N_2/N_1$$ | Voltage transformation | Ideal transformer |
| Current ratio: $$I_2/I_1 = N_1/N_2$$ | Conservation of power | Ideal, 100 % efficiency |
| Power: $$E_1 I_1 = E_2 I_2$$ | Equality of input & output | Ideal |
| Efficiency: $$\eta = P_{\text{out}}/P_{\text{in}}$$ | Practical transformer | Include losses |
| AC ammeter reading: $$I_{\mathrm{rms}}$$ | Calibrated to RMS | Assumes sine wave |
| AC voltmeter reading: $$E_{\mathrm{rms}}$$ | — | — |
Worked Example
An ideal transformer has 500 primary turns and 50 secondary turns. The primary is connected to 220 V (rms) mains and draws 0.5 A. Find (a) secondary voltage, (b) secondary current.
- Voltage ratio $$E_2 = E_1 (N_2/N_1) = 220 \times (50/500) = 22\ \text{V}$$.
- Current ratio $$I_2 = I_1 (N_1/N_2) = 0.5 \times (500/50) = 5\ \text{A}$$.
Answer: (a) 22 V (rms), (b) 5 A (rms)
Shortcuts and Special Cases
- If turn ratio is 1:10, voltage steps down by 10 but current steps up by 10—the opposite sense.
Watch Out
- Using peak values in transformer formulas; always RMS.
- Ignoring copper and iron losses when the question says “practical transformer”.
Commonly Confused Alternating Currents Formulas for JEE
| Confusing Pair | One is for… | The other is for… | How to decide |
|---|---|---|---|
| $$\omega_0 = 1/\sqrt{LC}$$ vs $$\omega = 1/\sqrt{LC}$$ | Resonant angular frequency | Instantaneous angular velocity (rare) | Resonance question will mention “at resonance” or equal reactances. |
| $$P = EI\cos\phi$$ vs $$Q = EI\sin\phi$$ | Average (real) power | Reactive power | If wording says “wattless” or “VArs”, use sine. |
| Series $$Z= \sqrt{R^2 + (X_L-X_C)^2}$$ | Series LCR | Parallel uses admittance | Check whether elements share same current (series) or voltage (parallel). |
| $$Q = \omega_0 L / R$$ vs $$Q = 1/(R\omega_0 C)$$ | Inductive form | Capacitive form | Pick the one that avoids dividing by a tiny number. |
Common Mistakes to Avoid When Using Alternating Currents Formulas for JEE
- Substituting peak instead of RMS values in power and transformer questions.
- Forgetting to convert milli-, micro-, and kilo- prefixes to SI before plugging into $$\omega_0 = 1/\sqrt{LC}$$.
- Adding reactances arithmetically instead of vectorially when elements are in series.
- Writing $$X_L = 1/(\omega L)$$—inverse of correct formula.
- Assuming power in pure L or C is non-zero; it is zero on average.
- Using degree mode throughout and then mixing radians for $$\omega t$$ substitutions.
- Interchanging max/min characteristics of series and parallel resonance.
- Quoting $$P = EI$$ without the power factor; valid only when $$\phi=0$$.
Quick Revision of Alternating Currents Formulas for JEE Main and Advanced
- $$I_{\mathrm{rms}} = I_0/\sqrt2$$ and $$E_{\mathrm{rms}} = E_0/\sqrt2$$ – peak-to-RMS.
- $$X_L = \omega L$$, $$X_C = 1/\omega C$$ – reactances.
- Series LCR impedance $$Z=\sqrt{R^2 + (X_L - X_C)^2}$$ – one-line life-saver.
- Resonance $$\omega_0 = 1/\sqrt{LC}$$ – frequency at which $$X_L = X_C$$.
- Average power $$P = EI\cos\phi$$ – include the cos.
- Quality factor $$Q = \omega_0 L/R = 1/(R\omega_0 C)$$ – sharpness metric.
- Transformer ratio $$E_2/E_1 = N_2/N_1$$ – volts proportional to turns.
- Voltage magnification at resonance $$V_L/E = Q$$ – survey favourite.
Alternating Currents Formulas for JEE 2027: Conclusion
Alternating Currents formulas for JEE are easier to apply when you understand RMS values, phase relations and the behaviour of resistors, inductors and capacitors. Use this formula sheet to revise AC circuit impedance, LCR resonance, power factor and transformer relations. Pay attention to the conditions behind each formula, especially whether the circuit is series or parallel and whether the given values are peak or RMS.
Follow each revision session with JEE Alternating Currents PYQs and questions from JEE Physics mock tests. Check your units, draw phasor diagrams where helpful and review mistakes before moving to harder problems. Combining formula revision with regular practice can improve your speed and accuracy in Alternating Current questions for JEE Main and Advanced.
