RL Circuits — Current growth and decay
EMI
7
JEE Qs
8%
Hard
75
min
Master the initial (t=0) and final (t=infinity) conditions for inductors to quickly solve problems or verify your calculations for transient states.
🧮 Key Formulas
✅ Key Points for JEE
- 1At t=0 (just after closing the switch or connecting the source), an inductor behaves like an open circuit, meaning the current through it is zero or maintains its initial value.
- 2At t=infinity (steady state), an inductor behaves like a short circuit, meaning the voltage across it is zero and the current through it becomes constant.
- 3The time constant (τ = L/R) dictates how quickly the current reaches its steady-state value (growth) or decays to zero (decay). After one time constant, current changes by approximately 63.2% of its total change.
- 4Always apply Kirchhoff's Voltage Law (KVL) to derive the differential equation for the circuit and then solve it, or directly apply the growth/decay formulas with correct initial and final conditions.
- 5For decay, the initial current (I_0) is the steady-state current just before the circuit change that initiates decay.
⚠️ Common Mistakes
- ✕Incorrectly identifying the initial current (I_0) or steady-state current (I_max) at t=0 and t=infinity, especially in complex circuits or when multiple switches are involved.
- ✕Confusing the formulas for current growth and current decay, or applying them with incorrect signs or constants.
- ✕Calculating the effective resistance (R_effective) incorrectly, particularly when parts of the circuit are shorted or opened, or when there are multiple resistors.
- ✕Forgetting that the energy stored in the inductor is dissipated as heat in the resistor during current decay, often leading to questions about energy conservation.
📝 Practice Questions
See allQ42.Regarding self-inductance: A. The self-inductance of the coil depends on its geometry. B. Self-inductance does not depend on the permeability of the medium. C. Self-induced e.m.f. opposes any change in the current in a circuit. D. Self-inductance is electromagnetic analogue of mass in mechanics. E. Work needs to be done against self-induced e.m.f. in establishing the current. Choose the correct answer from the options given below: (1) A, B, C, E only (2) B, C, D, E only (3) A, C, D, E only (4) A, B, C, D only
Q49. A conducting bar moves on two conducting rails as shown in the figure. A constant magnetic field B exists into the page. The bar starts to move from the vertex at time t = 0 with a constant velocity. If the induced EMF is E ∝tn , then value of n is _.
Q35.A rectangular metallic loop is moving out of a uniform magnetic field region to a field free region with a constant speed. When the loop is partially inside the magnate field, the plot of magnitude of induced emf (ε) with time (t) is given by 2025 (22 Jan Shift 2) JEE Main Previous Year Paper (1) (2) (3) (4)
Q48.A parallel plate capacitor of area A = 16 cm2 and separation between the plates 10 cm , is charged by a DC current. Consider a hypothetical plane surface of area A0 = 3.2 cm2 inside the capacitor and parallel to the plates. At an instant, the current through the circuit is 6A. At the same instant the displacement current through A0 is ________ mA .
Q31.A uniform magnetic field of 0.4 T acts perpendicular to a circular copper disc 20 cm in radius. The disc is having a uniform angular velocity of 10πrads−1 about an axis through its centre and perpendicular to the disc. What is the potential difference developed between the axis of the disc and the rim ? (π = 3.14) (1) 0.5024 V (2) V (3) 0.2512V V (4) 0.1256V V
Q27.A coil of area A and N turns is rotating with angular velocity ω in a uniform magnetic field →B about an axis perpendicular to →B. Magnetic flux φ and induced emf ε across it, at an instant when →B is parallel to the plane of coil, are : (1) φ = AB, ε = 0 (2) φ = 0, ε = 0 (3) φ = 0, ε = NABω (4) φ = AB, ε = NABω
NCERT Chapters
- Class 12 Physics Ch 3: Current Electricity
- Class 12 Physics Ch 6: Electromagnetic Induction