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

Given:
Inductance of the inductor, L = 500 mH
Resistance of the resistor connected, R = 25 Ω
Emf of the battery, E = 5 V

For the given circuit, the potential difference across the resistance is given by
V = iR

The current in the LR circuit at time t is given by
i = i₀ (1 − e^−tR/L)
∴ Potential difference across the resistance at time t, V = (i₀(1 − e^−tR/L)R

(a) For t = 20 ms,
i = i₀(1 − e^−tR/L)
$$\begin{gathered} =\frac{E}{R}(1-e^{-tR/L}) \\ =\frac{5}{25}(1-e^{-(2\times {10}^{-3}\times 25)/(500\times {10}^{-3})} \\ =\frac{1}{5}(1-e^{-1})=\frac{1}{5}(1-0.3678) \\ =\frac{0.632}{5}=0.1264\mathrm{A} \end{gathered}$$
Potential difference:
V = iR = (0.1264) × (25)
= 3.1606 V = 3.16 V

(b) For t = 100 ms,
i = i₀(1 − e^−tR/L)
$$\begin{gathered} =\frac{5}{25}\left(1-e^{(-100\times {10}^{-3})\times (25/500\times {10}^{-3})}\right) \\ =\frac{1}{5}(1-e^{-50}) \\ =\frac{1}{5}(1-0.0067) \\ =\frac{0.9932}{5}=0.19864\mathrm{A} \end{gathered}$$
Potential difference:
V = iR
= (0.19864) × (25) = 4.9665 = 4.97 V

(c) For t = 1 s,
$$\begin{gathered} i=\frac{5}{25}\left(1-e^{-1\times 25/500\times {10}^{-3}}\right) \\ =\frac{1}{5}(1-e^{-50}) \\ =\frac{1}{5}\times 1=\frac{1}{5}\mathrm{A} \end{gathered}$$
Potential difference:
V = iR
$=\left(\frac{1}{5}\times 25\right)=5\mathrm{V}$