LR circuit — growth and decay of current, time constant

Question

How does current grow and decay in an LR circuit? What is the LR time constant and how does it compare with the RC circuit?

Solution — Step by Step

An inductor $L$ in series with resistor $R$ , connected to a battery of emf $\varepsilon$ :

The inductor opposes sudden changes in current (Lenz’s law). Current grows gradually:

i(t) = \frac{\varepsilon}{R}\left(1 - e^{-Rt/L}\right) = I_0\left(1 - e^{-t/\tau}\right)

where $I_0 = \varepsilon/R$ is the final (steady-state) current and $\tau = L/R$ is the time constant.

At $t = 0$ : current is zero (inductor blocks sudden change). At $t = \infty$ : current reaches $\varepsilon/R$ (inductor acts as a plain wire).

When the battery is removed and the circuit is closed through just $R$ and $L$ :

i(t) = I_0 \, e^{-Rt/L} = I_0 \, e^{-t/\tau}

The current decays exponentially. After one time constant, it drops to $I_0/e \approx 37\%$ of its initial value.

The inductor now acts as a source of emf, trying to maintain the current that was flowing.

\tau = \frac{L}{R}

Compare with RC circuit where $\tau = RC$ .

Property	RC Circuit	LR Circuit
Time constant	$\tau = RC$	$\tau = L/R$
Quantity that grows/decays	Charge $q$	Current $i$
Initial state (growth)	$q = 0$ , $i = \varepsilon/R$	$i = 0$ , $V_L = \varepsilon$
Final state (growth)	$q = C\varepsilon$ , $i = 0$	$i = \varepsilon/R$ , $V_L = 0$
Larger R means	Slower process	Faster process

Notice the opposite effect of $R$ : in RC circuits, larger $R$ means slower charging ( $\tau = RC$ increases). In LR circuits, larger $R$ means faster current growth ( $\tau = L/R$ decreases). This is a favourite comparison question in JEE.

At steady state, the energy stored in the inductor’s magnetic field:

U = \frac{1}{2}LI_0^2 = \frac{L\varepsilon^2}{2R^2}

During current decay, this energy is entirely dissipated as heat in the resistor.

flowchart TD
    A["LR Circuit Problem"] --> B{"Growth or decay?"}
    B -->|"Growth: battery connected"| C["i = emf/R times 1 minus exp of minus Rt/L"]
    B -->|"Decay: battery removed"| D["i = I0 times exp of minus Rt/L"]
    A --> E["Time constant tau = L/R"]
    E --> F["At t = tau: 63% of final current reached"]
    E --> G["At t = 5tau: practically at steady state"]
    A --> H["Energy stored: U = half L I squared"]

Why This Works

Applying KVL to the LR growth circuit:

\varepsilon = iR + L\frac{di}{dt}

This is analogous to the RC equation but with current instead of charge. The inductor’s back-emf ( $L \, di/dt$ ) opposes the change in current, just as the capacitor’s voltage opposes further charge accumulation. The solution is the same exponential form with $\tau = L/R$ .

Alternative Method

For quick analysis without solving the ODE, use the initial and final conditions approach. At $t = 0^+$ , the inductor acts as an open circuit (blocks current). At $t = \infty$ , the inductor acts as a short circuit (zero resistance for DC). Knowing these two limits plus the exponential nature gives you the complete solution.

Common Mistake

When the battery is suddenly disconnected from an LR circuit, students assume the current drops to zero instantly. It does not — the inductor generates an emf to maintain current flow. If there is no alternate path for current (open circuit), the inductor generates a very large voltage spike (theoretically infinite). In practice, this causes sparking at switches. In problems, always check that a closed loop exists for current decay. JEE Advanced tests this concept in circuit-switching problems.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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