Question

An LR circuit contains an inductor of 500 mH, a resistor of 25.0 Ω and an emf of 5.00 V in series. Find the potential difference across the resistor at t = (a) 20.0 ms, (b) 100 ms and (c) 1.00 s.

Answer 1

Answer:

100 ms is the answer of your questions

Answer 2

The potential difference across the resistor at t = 20.0 ms is 3.16 V

The potential difference across the resistor at t = 100 ms is 4.97 V

The potential difference across the resistor at t =  1.00 s is 5 V

Explanation:

The inductive inductance, L = 500 mH

Resistance of the linked resistor, R = 25 Ω

Emf battery, E = 5 V

The potential difference over the resistance for the given circuit is given by

$V=i R$

The current at time t in the LR circuit is indicated by

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

∴ Potential gap in time t over the resistance, $\mathrm{v}=i_{0}\left(1-e^{\frac{-R}{L}}\right) \mathrm{R}$

(a) For t = 20 ms,  

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

 $=\frac{E}{R}\left(1-e^{\frac{-t R}{L}}\right)$

 $=\frac{5}{25}\left(1-e^{\frac{-2 \times 10^{3} \times 25}{500 \times 10^{-3}}}\right)$

 $=15\left(1-e^{-1}\right)=\frac{1}{5}(1-0.3678)$

$=\frac{0.632}{5}=0.1264 A$

The potential difference across the resistor

V = iR

  $=(0.1264) \times(25)$

  = 3.1606 V

  = 3.16 V

(b) at time t = 100 ms,

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

 $=\frac{5}{25}\left(1-e^{\frac{-100 \times 10^{-3}}{25 \times 10^{-3}}}\right)$

 $= \frac{1}{5} \left(1-e^{-50}\right)$

 $= \frac{1}{5} (1-0.0067)$

 = 0.9932/5

 = 0.19864 A

The potential difference across the resistor

V = iR

  = (0.19864) × (25)

  = 4.9665

  = 4.97 V

(c) at time t = 1 s,

  $=\frac{5}{25}\left(1-e^{\frac{-1 \times 25}{500 \times 10^{-3}}}\right)$

  $=\frac{1}{5}\left(1-e^{-50}\right)$

  $=\frac{1}{5}(1)$

  $=\frac{1}{5} A$

The potential difference across the resistor

V = iR

  $=\frac{1}{5} \times 25$

  $=5 V$