---
title: "Time for current to reach 80 percent"
subject: "Engineering Physics"
chapter: 8
key: "semester_1/engineering_physics:solenoid-lr-eighty-percent-current"
source: https://www.bctengineer.com/q/semester_1/engineering_physics/solenoid-lr-eighty-percent-current
site: BCT Engineering
---

# Time for current to reach 80 percent

Engineering Physics · Chapter 8: Electromagnetism

- Asked in: 2072 Kartik, 5 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/engineering_physics/solenoid-lr-eighty-percent-current
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-8

## Question

A solenoid having an inductance of 6.3 μH is connected in series with a 1.2 kΩ resistance. If a 14 V battery is connected across the pair, how long will it take for the current through the resistor to reach 80% of its final value?

## Short answer

Time constant τ = L/R = 5.25 ns. Time to reach 80% of final current is t = τ ln(1/0.2) ≈ 8.45 ns.

## Answer

For an LR circuit connected to a battery, the current grows as I = I_max(1 − e^(−t/τ)), where τ = L/R and I_max = ε/R.

**Given:** L = 6.3 μH, R = 1.2 kΩ, ε = 14 V.

```text
τ     = L/R = 6.3×10⁻⁶ / 1.2×10³ = 5.25×10⁻⁹ s
I_max = ε/R = 14 / 1200 = 1.167×10⁻² A = 11.67 mA

I = 0.80 I_max  →  1 − e^(−t/τ) = 0.80
                   e^(−t/τ)     = 0.20
                   t/τ = ln 5   = 1.609
t = 1.609 × 5.25×10⁻⁹ = 8.45×10⁻⁹ s
```

At that time the current is 0.8 × 11.67 mA = 9.33 mA.

**Answer: t ≈ 8.45 ns (about 1.61 time constants)**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. The time constant of an LR circuit is:
   A. τ = R/L
   B. τ = L/R
   C. τ = LR
   D. τ = 1/(LR)
   Answer: B. Here τ = 6.3 μH/1.2 kΩ = 5.25 ns. (Key point: Time constant)

2. How does the current grow after a battery is connected to a series LR circuit?
   A. I = I_max(1 − e^(−t/τ)), with I_max = ε/R
   B. I = I_max e^(−t/τ)
   C. I = I_max(1 − e^(−t/τ)), with I_max = εL
   D. I = I_max t/τ until it reaches I_max
   Answer: A. The inductor opposes the rise, so current grows exponentially toward ε/R. (Key point: Growth equation)

3. Setting I = 0.80 I_max gives t/τ equal to:
   A. ln 0.8 ≈ −0.22
   B. ln 2 ≈ 0.69
   C. ln 5 ≈ 1.61
   D. 0.80
   Answer: C. 1 − e^(−t/τ) = 0.8 so e^(−t/τ) = 0.2 and t/τ = ln 5. (Key point: Solving for 80%)

4. The time for the current to reach 80% of its final value is about:
   A. 8.45 μs
   B. 8.45 ns
   C. 5.25 ns
   D. 1.61 ns
   Answer: B. t = 1.609 × 5.25 ns ≈ 8.45 ns. (Key point: Final time)

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