Topic 07 of 14

LR circuits: growth, decay and time constant

Close a switch on a coil and the current creeps up instead of jumping. This file times that creep with one number, τ = L/R, and gives quick pictures for the instant after switching and long after.

Beyond NCERTAllen module pages 111–115Illustrations 19–20, Beginner's Box 5

Switching on: the current grows

Growth and decay. The current does not jump. It climbs quickly at first and then more and more slowly towards E/R. When the battery is taken out and the circuit shorted, it dies away the same exponential way.

Picture it

Push a heavy boat through water with a steady force. At first it barely moves, but it speeds up. The faster it goes, the more the water drags on it, so it gains speed more and more slowly until it cruises at a top speed. The battery is the steady push, the resistor is the water drag, and the inductor is the boat's heaviness. The current behaves exactly like the boat's speed.

In exam language

Kirchhoff's loop rule for a cell E, inductor L and resistance R in series (R includes the coil's own resistance):

E − L dI/dt − IR = 0 I = I₀(1 − e^−t/τ), I₀ = E/R, τ = L/R

Just after the switch is closed (t = 0) the current is zero, so the inductor acts like a break in the circuit. Long after, dI/dt = 0, so the inductor acts like a plain wire.

Trap

The final current E/R does not depend on L. The inductor changes only how long it takes to get there.

Time constant and half-life

The time constant τ = L/R has the unit of time (henry ÷ ohm = second). It sets the clock for the whole process.

Time after closing	Current as a fraction of I₀
τ	1 − e⁻¹ = 0.632 (63%)
2τ	0.865 (86%)
3τ	0.950 (95%)
5τ	0.993 (practically steady)

The half-life is the time taken to reach half of I₀ (or, during decay, to fall to half):

1 − e^−t/τ = ½ → t_½ = τ ln 2 = 0.693 L/R

Rate of growth

dI/dt = (E/L) e^−t/τ largest at t = 0: (dI/dt)_max = E/L

The starting rate E/L does not depend on R, because at t = 0 there is no current and so no voltage across R: all of E sits across the inductor.

Sharing E. At every instant V_L + V_R = E. The inductor takes it all at the start and none at the end.

Switching off: the current decays

If the battery is simply disconnected, the current has nowhere to go and the inductor produces a huge emf that sparks across the switch. So the module uses a make-before-break switch: it first connects a wire across L and R, then removes the battery. The current then decays through that path:

L dI/dt + IR = 0 → I = I₀ e^−t/τ

Quantity	Growth	Decay
Current	I₀(1 − e^−t/τ)	I₀ e^−t/τ
At t = τ	0.63 I₀	0.37 I₀
Half-life	0.693 τ	0.693 τ
\|dI/dt\|	(E/L)e^−t/τ	(I₀/τ)e^−t/τ

During decay, all the energy ½LI₀² that the inductor stored turns into heat in R.

Switching snapshots: just after and long after

Two bulbs. The coil delays the current in its own branch, so bulb P lags. Once steady, the coil is just a resistance equal to R, so the bulbs match.

For circuits with several branches, two quick pictures replace all the exponentials:

Moment	Replace the inductor by	Why
Just after closing (t = 0⁺)	An open gap (no current through it)	Its current cannot jump from zero
Long after closing (t → ∞)	A plain wire (its own resistance only)	dI/dt = 0, so no emf
Just after opening	A current source carrying its old current	Its current cannot jump to zero

Allen Illustration 19. A 10 V cell, 2 Ω in series, then a 3 Ω branch in parallel with a branch of 6 Ω and L. Just after closing, the L-branch is open: I = 10/(2 + 3) = 2 A. Long after, 3 Ω ∥ 6 Ω = 2 Ω: I = 10/(2 + 2) = 2.5 A. Checked: correct.

Combining inductors

When coils are far enough apart that their fields do not link each other, inductors combine like resistors:

series: L = L₁ + L₂ parallel: 1/L = 1/L₁ + 1/L₂

Allen notes that cutting a coil into two equal halves leaves the time constant unchanged: each half has half the inductance (same turns per metre, half the length) and half the resistance, so L/R is the same.

Formula sheet

Formula	Meaning	When to use	Watch out
I = I₀(1 − e^−t/τ)	Growth	Switch closed on L, R, E	I₀ = E/R, not E/L
I = I₀ e^−t/τ	Decay	Battery removed, circuit shorted	37% after τ
τ = L/R	Time constant	Every LR question	R is the total resistance in the loop
t_½ = 0.693 L/R	Half-life	Time to 50%	ln 2, not ½
dI/dt = (E/L)e^−t/τ	Rate of growth	Rate at any time	Initial rate E/L is independent of R
V_L = E e^−t/τ	Voltage across L during growth	Voltage sharing	V_L + V_R = E
t = τ ln[I₀/(I₀ − I)]	Time to reach a current I	Given a target current	Growth only
L = L₁ + L₂; 1/L = 1/L₁ + 1/L₂	Combinations	No mutual coupling	Like resistors

Allen pages 111–115 checked

Item	Check
Growth I = I₀(1 − e^−t/τ), τ = L/R	Correct
Peak current independent of L; initial rate independent of R	Correct
Time constant: 63% (growth), 37% (decay); half-life 0.693 L/R	Correct
Decay with make-before-break switch	Correct
Series and parallel combination	Correct (coils not coupled)
Cutting a coil into two parts keeps τ the same	Correct
Illustration 19: 2 A and 2.5 A	Correct
Illustration 20: Q first; equal later	Correct
Box 5 Q8 option (1) "Change"	Misprint for "Charge"

Beginner's Box 5 answer key

Q	Answer	Working
1	(4) C/L	1/RC, R/L and 1/√(LC) are all 1/time. C/L is not.
2	(2) 1 − e⁻¹	I₀ = 5/5 = 1 A, τ = 10/5 = 2 s. At t = 2 s: I = 1 − e⁻¹.
3	(1) 2 ms	τ = L/R = 8 mH/4 Ω = 2 ms.
4	(1) 0.63 I₀	Definition of the time constant during growth.
5	(4) 0.25 A/s	Initial rate = E/L = 5/20.
6	(1) 27.3 A/s	dI/dt = (E/L)e^−Rt/L = (5/5 × 10⁻³) e^{−18 × 0.001/0.005} = 1000 × e^−3.6 = 27.3 A/s.
7	(4) 1 ms	I₀ = 12/6 = 2 A; 1 A is half, so t = 0.693 × (8.4 × 10⁻³/6) = 9.7 × 10⁻⁴ s ≈ 1 ms.
8	(4) Current⁻¹	L/R is a time and CV is a charge, so L/(CVR) = time/charge = 1/current.
9	(2) 3.3 A, 3.3 A, 0	Just after closing the L branch carries no current. i₁ = i₂ = 10/(1 + 2) = 3.3 A, i₃ = 0.

NEET practice: 30 questions

The exponential forms are Beyond NCERT, but the switching snapshots (just after, long after) are asked as circuit reasoning. Every number was recalculated in Python.

Q1Numerical

An LR circuit has L = 40 mH and R = 8 Ω. Its time constant is

(A)320 ms
(B)5 ms
(C)0.2 s
(D)5 s

Show the solution

Given

L = 0.04 H, R = 8 Ω

Asked

Concept

Time constant of an LR circuit.

Formula

τ = L/R

Baby steps

τ = 0.04/8 = 0.005 s = 5 ms.

Answer

(B) 5 ms

Why not the others

320 ms multiplies L and R. 0.2 s is R/L. 5 s forgets that L is in millihenry.

Shortcut

Henry ÷ ohm = seconds.

Where it went wrong

Inverting the ratio.

Q2Concept

A 12 V cell is connected through a switch to an inductor and a 4 Ω resistor in series. If the inductance is doubled, the final steady current

(A)becomes 1.5 A
(B)becomes 6 A
(C)stays 3 A
(D)becomes zero

Show the solution

Given

E = 12 V, R = 4 Ω, L doubled

Asked

Final current

Concept

At steady state the inductor has no emf.

Formula

I₀ = E/R

Baby steps

I₀ = 12/4 = 3 A.
L does not appear, so it stays 3 A.

Answer

Why not the others

6 A and 1.5 A make I₀ depend on L. Zero ignores the steady state.

Shortcut

L sets the timing, not the final value.

Where it went wrong

Putting L into the steady-state current.

Q3Numerical

In an LR circuit the steady current is 2 A. The current one time constant after the switch is closed is about

(A)1.73 A
(B)0.74 A
(C)1.0 A
(D)1.26 A

Show the solution

Given

I₀ = 2 A, t = τ

Asked

Concept

Growth reaches 63% of I₀ in one τ.

Formula

I = I₀(1 − e⁻¹)

Baby steps

1 − e⁻¹ = 0.632.
I = 2 × 0.632 = 1.26 A.

Answer

(D) 1.26 A

Why not the others

0.74 A is the 37% value (decay). 1.0 A is half (the half-life value). 1.73 A is the 2τ value.

Shortcut

63% of I₀.

Where it went wrong

Using the decay fraction for growth.

Q4Numerical

A 4 H coil of resistance 2 Ω is connected to a 10 V cell at t = 0. The current at t = 2 s is about

(A)5.0 A
(B)1.84 A
(C)3.16 A
(D)4.32 A

Show the solution

Given

L = 4 H, R = 2 Ω, E = 10 V, t = 2 s

Asked

Concept

Growth.

Formula

I = (E/R)(1 − e^−Rt/L)

Baby steps

I₀ = 5 A, τ = 2 s.
t/τ = 1.
I = 5(1 − 0.368) = 3.16 A.

Answer

Why not the others

1.84 A is 5e⁻¹ (decay form). 5.0 A is the final current. 4.32 A uses t/τ = 2.

Shortcut

t = τ → 63% of 5 A.

Where it went wrong

Using the decay formula.

Q5Numerical

A 20 V cell is switched on across a 5 H inductor in series with a 10 Ω resistor. The initial rate of growth of current is

(A)4 A/s
(B)2 A/s
(C)0.25 A/s
(D)40 A/s

Show the solution

Given

E = 20 V, L = 5 H, R = 10 Ω

Asked

(dI/dt) at t = 0

Concept

At t = 0, I = 0, so all of E is across L.

Formula

(dI/dt)₀ = E/L

Baby steps

E/L = 20/5 = 4 A/s.

Answer

(A) 4 A/s

Why not the others

2 A/s is E/R (a current). 0.25 A/s inverts E/L. 40 A/s multiplies E by R/L.

Shortcut

R drops out at t = 0.

Where it went wrong

Using E/R.

Q6Numerical

For a 10 V cell and a 2 H inductor in an LR circuit, the rate of growth of current one time constant after switching on is about

(A)1.84 A/s
(B)5.0 A/s
(C)3.16 A/s
(D)0.68 A/s

Show the solution

Given

E = 10 V, L = 2 H, t = τ

Asked

dI/dt

Concept

Rate of growth decays exponentially.

Formula

dI/dt = (E/L)e^−t/τ

Baby steps

E/L = 5 A/s.
× e⁻¹ = 5 × 0.368 = 1.84 A/s.

Answer

(A) 1.84 A/s

Why not the others

5.0 A/s is the initial rate. 3.16 A/s uses (1 − e⁻¹). 0.68 A/s uses e⁻² (two time constants).

Shortcut

Initial rate × 0.37.

Where it went wrong

Using the growth factor (1 − e⁻¹) for the rate.

Q7Numerical

An LR circuit has L = 0.3 H and R = 6 Ω. The time taken for the current to reach half its final value is about

(A)69 ms
(B)50 ms
(C)25 ms
(D)35 ms

Show the solution

Given

L = 0.3 H, R = 6 Ω

Asked

t_½

Concept

Half-life.

Formula

t_½ = 0.693 L/R

Baby steps

τ = 0.05 s.
t_½ = 0.693 × 0.05 = 0.0347 s ≈ 35 ms.

Answer

(D) 35 ms

Why not the others

50 ms is τ itself. 25 ms takes half of τ. 69 ms uses 2τ ln 2.

Shortcut

0.69 × τ.

Where it went wrong

Taking half of τ.

Q8Concept

In an LR circuit, the time taken for the current to grow from zero to 75% of its final value is

(A)0.75 τ
(B)τ ln 4
(C)τ ln(4/3)
(D)2τ

Show the solution

Given

I = 0.75 I₀

Asked

t in terms of τ

Concept

Solve the growth equation.

Formula

1 − e^−t/τ = 0.75

Baby steps

e^−t/τ = 0.25.
t/τ = ln 4.
t = τ ln 4 ≈ 1.39 τ.

Answer

(B) τ ln 4

Why not the others

0.75τ treats growth as linear. τ ln(4/3) solves e^−t/τ = 0.75. 2τ gives 86%.

Shortcut

Remaining fraction 1/4 → ln 4.

Where it went wrong

Setting e^−t/τ equal to the reached fraction instead of the remaining one.

Q9Numerical

A current of 4 A decays through an LR circuit with a time constant of 10 ms. The current after 20 ms is about

(A)1.47 A
(B)0.54 A
(C)3.46 A
(D)1.0 A

Show the solution

Given

I₀ = 4 A, τ = 10 ms, t = 20 ms

Asked

Concept

Decay.

Formula

I = I₀ e^−t/τ

Baby steps

t/τ = 2.
I = 4e⁻² = 4 × 0.135 = 0.54 A.

Answer

(B) 0.54 A

Why not the others

1.47 A is one τ. 3.46 A uses the growth formula. 1.0 A halves twice (two half-lives, not two τ).

Shortcut

Two time constants → × 0.135.

Where it went wrong

Confusing time constants with half-lives.

Q10Numerical

A 0.2 H inductor carrying 3 A is shorted through a resistor, and the current decays to zero. The total heat produced in the resistor is

(A)1.8 J
(B)0.6 J
(C)0.9 J
(D)0.3 J

Show the solution

Given

L = 0.2 H, I₀ = 3 A

Asked

Heat

Concept

All the stored energy becomes heat.

Formula

H = ½LI₀²

Baby steps

½ × 0.2 × 9 = 0.9 J.

Answer

Why not the others

0.6 J is LI. 1.8 J forgets the ½. 0.3 J uses I instead of I².

Shortcut

Stored energy = heat released.

Where it went wrong

Trying to integrate I²R with an unknown R.

Q11Concept

One time constant after an LR circuit is switched on, the energy stored in the inductor is what fraction of its final value?

(A)about 0.86
(B)about 0.63
(C)about 0.37
(D)about 0.40

Show the solution

Given

I = 0.632 I₀ at t = τ

Asked

U/U_final

Concept

Energy goes as I².

Formula

U ∝ I²

Baby steps

I/I₀ = 0.632.
U/U_final = 0.632² = 0.40.

Answer

(D) about 0.40

Why not the others

0.63 is the current fraction. 0.37 is the decay fraction. 0.86 is the current at 2τ.

Shortcut

Square the current fraction.

Where it went wrong

Forgetting to square.

Q12Numerical

A 12 V cell is switched on across an inductor and resistor in series. The voltage across the inductor one time constant later is about

(A)7.6 V
(B)4.4 V
(C)6.0 V
(D)12 V

Show the solution

Given

E = 12 V, t = τ

Asked

V_L

Concept

V_L falls exponentially from E.

Formula

V_L = E e^−t/τ

Baby steps

12 × 0.368 = 4.4 V.

Answer

(B) 4.4 V

Why not the others

7.6 V is across R. 6.0 V is the half-life value. 12 V is the value at t = 0.

Shortcut

Inductor's share shrinks; resistor's share grows.

Where it went wrong

Giving the resistor's voltage.

Q13Numerical

A 2 H inductor and a 10 Ω resistor are switched on across a cell. The voltages across L and R become equal after about

(A)0.14 s
(B)0.20 s
(C)0.10 s
(D)0.28 s

Show the solution

Given

L = 2 H, R = 10 Ω

Asked

t for V_L = V_R

Concept

Equal shares means each is E/2.

Formula

e^−t/τ = ½

Baby steps

τ = 0.2 s.
t = τ ln 2 = 0.139 s.

Answer

(A) 0.14 s

Why not the others

0.20 s is τ. 0.10 s halves τ. 0.28 s doubles the half-life.

Shortcut

Equal voltages happen at the half-life.

Where it went wrong

Answering τ.

Q14Numerical

Inductors of 6 H and 3 H, far enough apart not to affect each other, are connected in parallel. The equivalent inductance is

(A)9 H
(B)2 H
(C)4.5 H
(D)18 H

Show the solution

Given

L₁ = 6 H, L₂ = 3 H, parallel, no coupling

Asked

Concept

Parallel inductors combine like parallel resistors.

Formula

1/L = 1/L₁ + 1/L₂

Baby steps

1/L = 1/6 + 1/3 = 1/2.
L = 2 H.

Answer

(B) 2 H

Why not the others

9 H is the series value. 4.5 H is the average. 18 H multiplies.

Shortcut

Product over sum: 18/9.

Where it went wrong

Adding in parallel.

Q15Numerical

A 12 V cell with 2 Ω in series feeds two parallel branches: a 6 Ω resistor, and a 3 Ω resistor in series with an inductor. The cell currents just after closing the switch and long after are

(A)2.4 A and 3 A
(B)3 A and 1.5 A
(C)1.5 A and 1.5 A
(D)1.5 A and 3 A

Show the solution

Given

E = 12 V, 2 Ω in series; 6 Ω ∥ (3 Ω + L)

Asked

I(0⁺) and I(∞)

Concept

Inductor: open at t = 0⁺, wire at t → ∞.

Formula

I = E/R_total

Baby steps

Just after: L-branch open, R = 2 + 6 = 8 Ω → 1.5 A.
Long after: 6 ∥ 3 = 2 Ω, R = 4 Ω → 3 A.

Answer

(D) 1.5 A and 3 A

Why not the others

3 A and 1.5 A swaps the two moments. 1.5 A and 1.5 A ignores the inductor opening up. 2.4 A uses 2 + 3 at t = 0.

Shortcut

Just after: delete the L branch. Long after: L is a wire.

Where it went wrong

Treating the inductor as a wire at t = 0.

Q16Concept

Which of these does not have the dimensions of time?

(A)√(LC)
(B)L/R
(C)RC
(D)R/L

Show the solution

Given

Combinations of L, C, R

Asked

Not a time

Concept

L/R, RC and √(LC) are all time constants.

Formula

[L/R] = [RC] = [√(LC)] = T

Baby steps

L/R is a time.
RC is a time.
√(LC) is a time (1/ω).
R/L is 1/time.

Answer

(D) R/L

Why not the others

Each of the other three is a standard time constant.

Shortcut

R/L is the inverse of the LR time constant.

Where it went wrong

Mixing up L/R and R/L.

Q17Concept

In an LR circuit the resistance is doubled, with L and E unchanged. Then

(A)τ halves, I₀ halves, and the initial rate of growth is unchanged
(B)τ doubles, I₀ halves, and the initial rate halves
(C)τ halves, I₀ is unchanged, and the initial rate doubles
(D)τ is unchanged, I₀ halves, and the initial rate halves

Show the solution

Given

R → 2R

Asked

Effect on τ, I₀ and (dI/dt)₀

Concept

τ = L/R, I₀ = E/R, (dI/dt)₀ = E/L.

Formula

—

Baby steps

τ = L/2R: halves.
I₀ = E/2R: halves.
E/L has no R: unchanged.

Answer

(A) τ halves, I₀ halves, and the initial rate of growth is unchanged

Why not the others

Each other option gets at least one dependence wrong.

Shortcut

Initial rate: only E and L.

Where it went wrong

Letting R affect the initial rate.

Q18Concept

Just after the switch in an LR circuit is closed, the inductor behaves like

(A)an open circuit
(B)a short circuit
(C)a resistor of value R
(D)a capacitor

Show the solution

Given

t = 0⁺

Asked

Behaviour of L

Concept

Current through an inductor cannot change suddenly.

Formula

I(0⁺) = I(0⁻) = 0

Baby steps

Before closing, I = 0.
Just after, I is still 0 in that branch.
So it acts like a break.

Answer

(A) an open circuit

Why not the others

A short circuit is the long-time behaviour. It is not a resistor or a capacitor.

Shortcut

Just after: open. Long after: wire.

Where it went wrong

Swapping the two limits.

Q19Concept

In Allen Illustration 20, bulb P is in series with a coil and bulb Q with a resistor of equal resistance, both across the same cell. When the switch is closed

(A)both light together and stay equally bright
(B)P lights first; later both are equally bright
(C)Q lights first; later both are equally bright
(D)Q lights first; P stays dimmer forever

Show the solution

Given

Parallel branches: coil + P, resistor + Q

Asked

Order and final brightness

Concept

The coil delays the current in its branch; at steady state it is just a resistance.

Formula

I = I₀(1 − e^−t/τ) in the coil branch

Baby steps

Q's branch has no inductance: current appears at once.
P's branch current grows over a few τ.
Coil resistance = R, so the final currents are equal.

Answer

Why not the others

P first reverses the delay. 'Together' ignores the inductance. 'Dimmer forever' forgets the coil has the same resistance.

Shortcut

Coil = delay only.

Where it went wrong

Thinking an inductor permanently reduces current.

Q20Concept

A make-before-break switch is used when disconnecting the battery from an LR circuit so that

(A)the current stops instantly
(B)the current can decay through a closed path instead of producing a large spark
(C)the time constant becomes zero
(D)the energy stored is returned to the battery

Show the solution

Given

Removing the battery

Asked

Purpose of make-before-break

Concept

Stopping current in an inductor suddenly needs an enormous emf.

Formula

e = −L dI/dt

Baby steps

Breaking first would make dI/dt huge → huge emf → spark.
Making the new path first lets I decay smoothly as I₀e^−t/τ.

Answer

(B) the current can decay through a closed path instead of producing a large spark

Why not the others

Current in an inductor cannot stop instantly. τ is unchanged. The energy becomes heat in R, not charge in the battery.

Shortcut

Give the current somewhere to go.

Where it went wrong

Thinking the current can be switched off instantly.

Q21Graph

Which graph shows the current against time after an LR circuit is switched on?

Show the solution

Given

Growth

Asked

I against t

Concept

Exponential approach to E/R.

Formula

I = I₀(1 − e^−t/τ)

Baby steps

Starts at zero with the steepest slope.
Bends over and levels off at I₀.

Answer

(A) the graph in option A

Why not the others

The falling curve is decay. The ramp-then-flat graph rises at a constant rate and stops abruptly. The flat line is a plain resistor circuit.

Shortcut

Steep start, flat finish.

Where it went wrong

Drawing a straight ramp.

Q22Graph

Which graph shows the rate of change of current, dI/dt, against time after an LR circuit is switched on?

Show the solution

Given

Growth

Asked

dI/dt against t

Concept

Derivative of the growth curve.

Formula

dI/dt = (E/L)e^−t/τ

Baby steps

Maximum E/L at t = 0.
Falls exponentially towards zero.

Answer

Why not the others

The rising curve is I itself. A constant rate would give a straight-line current. A straight fall to zero is not exponential.

Shortcut

The slope of the growth curve shrinks exponentially.

Where it went wrong

Plotting I instead of dI/dt.

Q23Graph

Which graph shows the voltage across the inductor against time after an LR circuit is switched on?

Show the solution

Given

Growth

Asked

V_L against t

Concept

V_L = L dI/dt.

Formula

V_L = E e^−t/τ

Baby steps

At t = 0 all of E is across L.
It falls exponentially to zero.

Answer

(D) the graph in option D

Why not the others

The rising curve is V_R. Zero ignores the start. Constant E ignores the growth of current.

Shortcut

V_L starts at E and dies away.

Where it went wrong

Plotting the resistor's voltage.

Q24Graph

Two LR circuits have the same cell and the same resistance, but circuit 2 has the larger inductance. Which graph correctly compares their currents after switching on?

Show the solution

Given

Same E and R; L₂ > L₁

Asked

Comparison of growth curves

Concept

I₀ = E/R is the same; τ = L/R is larger for circuit 2.

Formula

τ ∝ L

Baby steps

Both curves level off at the same I₀.
Circuit 2 rises more slowly.

Answer

(B) the graph in option B

Why not the others

Different final currents would need different R. The graph where the slower curve also ends lower changes R too. A single curve ignores the difference in L.

Shortcut

Same destination, slower journey.

Where it went wrong

Letting L change the final current.

Q25Assertion–reason

Assertion (A): The final steady current in an LR circuit does not depend on the inductance.
Reason (R): At steady state there is no self-induced emf across the inductor.

(A)Both A and R are true, and R is the correct explanation of A.
(B)Both A and R are true, but R is not the correct explanation of A.
(C)A is true, but R is false.
(D)A is false, but R is true.

Show the solution

Given

A: I₀ independent of L. R: no emf at steady state.

Asked

Truth and link

Concept

Steady state: dI/dt = 0.

Formula

I₀ = E/R

Baby steps

A is true.
R is true.
With no emf across L, only R limits the current: R explains A.

Answer

(A) Both A and R are true, and R is the correct explanation of A.

Why not the others

(B) denies a direct link. (C) and (D) need a false statement.

Shortcut

No emf → L drops out.

Where it went wrong

Choosing (B) by habit.

Q26Assertion–reason

Assertion (A): The quantity L/R has the dimensions of time.
Reason (R): Inductance and resistance are both scalar quantities.

(A)Both A and R are true, and R is the correct explanation of A.
(B)Both A and R are true, but R is not the correct explanation of A.
(C)A is true, but R is false.
(D)A is false, but R is true.

Show the solution

Given

A: L/R is a time. R: both scalars.

Asked

Truth and link

Concept

Dimensions come from units, not from being scalar.

Formula

[L]/[R] = [T]

Baby steps

A is true.
R is true.
Being scalars says nothing about dimensions: R does not explain A.

Answer

(B) Both A and R are true, but R is not the correct explanation of A.

Why not the others

(A) needs a causal link. (C) and (D) need a false statement.

Shortcut

'A time because they are scalars' makes no sense.

Where it went wrong

Linking two true facts.

Q27Assertion–reason

Assertion (A): One time constant after an LR circuit is switched on, the current is 50% of its final value.
Reason (R): The time constant of an LR circuit is L/R.

(A)Both A and R are true, and R is the correct explanation of A.
(B)Both A and R are true, but R is not the correct explanation of A.
(C)A is true, but R is false.
(D)A is false, but R is true.

Show the solution

Given

A: 50% at τ. R: τ = L/R.

Asked

Truth and link

Concept

At τ the current is 63%; 50% is at the half-life.

Formula

1 − e⁻¹ = 0.63

Baby steps

R is true.
A is false.
A false, R true.

Answer

(D) A is false, but R is true.

Why not the others

(A), (B) and (C) need A true.

Shortcut

τ → 63%; 0.693τ → 50%.

Where it went wrong

Mixing up τ with the half-life.

Q28Assertion–reason

Assertion (A): Just after the switch in an LR circuit is closed, the current in it is zero.
Reason (R): Just after closing, the inductor behaves like a short circuit.

(A)Both A and R are true, and R is the correct explanation of A.
(B)Both A and R are true, but R is not the correct explanation of A.
(C)A is true, but R is false.
(D)A is false, but R is true.

Show the solution

Given

A: I(0⁺) = 0. R: L is a short at t = 0⁺.

Asked

Truth and link

Concept

At t = 0⁺ the inductor acts as an open circuit.

Formula

I(0⁺) = 0

Baby steps

A is true.
R is false: it behaves like an open circuit.
A true, R false.

Answer

Why not the others

(A) and (B) need R true. (D) needs A false.

Shortcut

Zero current ↔ open, not short.

Where it went wrong

Swapping the limits.

Q29Two statements

Statement I: The half-life of current growth in an LR circuit is 0.693 L/R.
Statement II: During decay, the current falls to about 37% of its initial value in one time constant.

(A)Both Statement I and Statement II are true.
(B)Both Statement I and Statement II are false.
(C)Statement I is true, but Statement II is false.
(D)Statement I is false, but Statement II is true.

Show the solution

Given

Half-life and decay at τ

Asked

Which are true

Concept

t_½ = τ ln 2; e⁻¹ = 0.37.

Formula

—

Baby steps

Statement I is true.
Statement II is true.

Answer

(A) Both Statement I and Statement II are true.

Why not the others

Every other option calls a true statement false.

Shortcut

Both are formula-sheet facts.

Where it went wrong

Doubting correct numbers.

Q30Two statements

Statement I: Inductors in series, with no mutual coupling, add like resistors in series.
Statement II: Inductors in parallel, with no mutual coupling, add directly: L = L₁ + L₂.

(A)Both Statement I and Statement II are true.
(B)Both Statement I and Statement II are false.
(C)Statement I is true, but Statement II is false.
(D)Statement I is false, but Statement II is true.

Show the solution

Given

Combination rules

Asked

Which are true

Concept

Inductors combine like resistors.

Formula

1/L = 1/L₁ + 1/L₂ in parallel

Baby steps

Statement I is true.
Statement II is false.

Answer

Why not the others

Both-true fails on II. Both-false fails on I. I-false-II-true reverses them.

Shortcut

Parallel: reciprocals.

Where it went wrong

Adding parallel inductors directly.

Answer key

1 B

2 C

3 D

4 C

5 A

6 A

7 D

8 B

9 B

10 C

11 D

12 B

13 A

14 B

15 D

16 D

17 A

18 A

19 C

20 B

21 A

22 C

23 D

24 B

25 A

26 B

27 D

28 C

29 A

30 C

Spread across letters: A 8, B 8, C 7, D 7. No letter repeats more than twice in a row. Question mix: Numerical 12, Concept 8, Graph 4, Assertion–reason 4, Two statements 2. Balancing seed 3.