Question

An inductor of inductance 2.00 H is joined in series with a resistor of resistance 200 Ω and a battery of emf 2.00 V. At t = 10 ms, find (a) the current in the circuit, (b) the power delivered by the battery, (c) the power dissipated in heating the resistor and (d) the rate at which energy is being stored in magnetic field.

Answer 1

Given:
Inductance of the inductor, L = 2 H
Resistance of the resistor connected to the inductor, R = 200 Ω,
Emf of the battery connected, E = 2 V

(a) The current in the LR circuit after t seconds after connecting the battery is given by
i = i₀(1 − e^−t/τ)
Here,
i₀ = Steady state value of current
i₀ = $\frac{E}{R}=\frac{2}{200}\mathrm{A}$
At time t = 10 ms, the current is given by
i $=\frac{2}{200}(1-e^{-10\times {10}^{-3}\times 200/2})$
i = 0.01(1 − e⁻¹)
i = 0.01(1 − 0.3678)
i = 0.01 × 0.632 = 6.3 mA

(b) The power delivered by the battery is given by
P = Vi
P = Ei₀(1 − e^−t/τ)
$$\begin{gathered} P=\frac{E^2}{R}(1-e^{-t/\tau }) \\ P=\frac{2\times 2}{200}(1-e^{10\times {10}^{-3}\times 200/2}) \end{gathered}$$
P = 0.02(1 − e⁻¹)
P = 0.01264 = 12.6 mW

(c) The power dissipated in the resistor is given by
P₁ = i²R
P₁ = [i₀(1 − e^−t/τ)]² R
P₁ = (6.3 mA)² × 200
P₁ = 6.3 × 6.3 × 200 × 10⁻⁵
P₁ = 79.38 × 10⁻⁴
P₁ = 7.938 × 10⁻³ = 8 mW

(d) The rate at which the energy is stored in the magnetic field can be calculated as:
W = $\frac{1}{2}L{i}^2$
W $=\frac{L}{2}{i_0}^2(1-e^{-t/\tau }{)}^2$
W = 2 × 10⁻²(0.225)
W = 0.455 × 10⁻²
W = 4.6 × 10⁻³
W = 4.6 mW