Question

An L-R circuit has L = 1.0 H and R = 20 Ω. It is connected across an emf of 2.0 V at t = 0. Find di/dt at (a) t = 100 ms, (b) t = 200 ms and (c) t = 1.0 s.

Answer 1

Answer:

An L-R circuit has L = 1.0 H and R = 20 Ω. It is connected across an emf of 2.0 V at t = 0. Find di/dt at (a) t = 100 ms, (b) t = 200 ms and (c) t = 1.0 so please tell me what you mean

Answer 2

(a) di/dt at  t = 100 ms is $0.27 \mathrm{A} / \mathrm{s}$

(b) di/dt at  t = 200 ms is $0.0366 \mathrm{A} / \mathrm{s}$

(c) di/dt at  t = 1.0 s is $41 \times 10^{-9}$

Explanation:

L = 1.0 H

R = 20 Ω

Emf of the battery = 2.0 V

L is Inductance,  

R is circuit resistance ,  

The time constant is

$\tau=\frac{L}{R}=\frac{1}{20}=0.05 \mathrm{s}$

Current of Steady-state is

$i_{0}=\frac{e}{R}=\frac{2}{20}=0.1 A$

Current for time t:

$i=i_{0}\left(1-e^{\frac{-t}{\tau}}\right)$

or

$i=\left(i_{0}-i_{0} e^{\frac{-t}{\tau}}\right)$

We get to differentiate between the two sides with respect to t

$\frac{d i}{d t}=-\left(i_{0} \times\left(-\frac{1}{\tau}\right) e^{\frac{-t}{\tau}}\right)$

   $=\frac{i_{0}}{\tau} e^{\frac{-t}{\tau}}$

(a) for time t = 100 ms,

$\frac{d i}{d t}=\frac{0.1}{0.05} \times e^{\frac{-0.1}{0.05}}$

   = 0.27 A/s

(b) for time  t = 200 ms,

$\frac{d i}{d t}=\frac{0.1}{0.05} \times e^{\frac{-0.2}{0.05}}$

   = 0.0366 A/s

(c) for time t = 1 s,

$\frac{d i}{d t}=\frac{0.1}{0.05} \times e^{-\frac{1}{0.05}}$    

   $=41 \times 10^{-9}$