We notice that the current through the induction coil changes with time.

Consequently, there is a change in potential difference across the induction coil.

We know that τ = L/R = 16 × 10^-3H / 20Ω = 0.8 × 10^-3 s

We can also write the exponent R/L at 1/τ for simplicity.

The potential difference across an inductor is given by:

ΔV_L = -LdI/dt, but I = I₀e^-t/τ

So, ΔV_L = -L(d/dt)(I₀e^-t/τ) = (-L)(-1/τ)I₀e^-t/τ = (L/τ)I₀e^-t/τ

i) At t = 0 ms, we have:

ΔV_L = [(16 × 10^-3 H)/(0.8 × 10^-3 s)](30 × 10^-3 A)e^{-(0 s)/(0.8 × 10-3 s)} = 0.60 V

The potential difference across the coil at t = 0 ms is 0.60 V.

ii) At t = 4 ms:

ΔV_L = [(16 × 10^-3 H)/(0.8 × 10^-3 s)](30 × 10^-3 A)e-(4 × 10^-3 s)/(0.8 × 10^-3 s) = 4.0 × 10^-3 V

The potential difference across the coil at t = 4 ms is 4.0 × 10^-3 V.

A 260 mH inductor is connected to a circuit, as shown. The change in current in the circuit, di/dt = constant. A voltmeter connected across points p and q reads 2.02 V, q lying on at lower potential. Does the current in the circuit rise or fall?