A 1 000-V battery, a 3 000-Ω resistor, and a 0.50-μF capacitor are connected in series with a switch. The time constant for such a circuit, designated by the Greek letter, τ, is defined as the time that the capacitor takes to charge to 63% of its capacity after the switch is closed. What is the current in the circuit at a time interval of τ seconds after the switch has been closed?

The current in the circuit at a time interval of τ seconds after the switch has been closed is 0.123 A

Explanation:

The time constant for an R and C in series circuit is given by τ = RC.

R = 3000 ohms, C = 0.5 × 10⁻⁶ F = 5.0 × 10⁻⁷ F

τ = 3000 × 5 × 10⁻⁷ = 0.015 s

The voltage across a capacitor as it charges is given be

V(t) = Vs (1 - e⁻ᵏᵗ)

where k = 1/τ

At the point when t = τ, the expassion becomes

V(t = τ) = 1000 (1 - e⁻¹) = 0.632 × 1000 = 632 V

Current flows as a result of potential difference,.

Current in the circuit at this time t =  τ is given by

I = (Vs - Vc)/R

Vs = source voltage = 1000 V

Vc = Voltage across the capacitor = 632 V

R = 3000 ohms

I = (1000 - 632)/3000 = 0.123 A

you have to draw the persons position over there

yup its correct its: c) b b