Answer:
a) 0.01232 A
b) 0.00075 s = 0.75 ms
c) 0.0045323 A = 4.532 mA
d) 3.894 V
Explanation:
R = 500 Ω
V = 6.16 V
C = 1.50 μF
Let Vs be the voltage of the emf source
Let Vc be the voltage across the capacitor at any time
a) Current flows as a result of potential difference between two points. So, the current flows according to difference in voltage between the emf source and the capacitor.
At time t = 0,
There is no voltage on the capacitor; Vc = 0 V
Current in the circuit is given by
I = (Vs - Vc)/R
I = (6.16 - 0)/500
I = 0.01232 A
b) Time constant for an RC circuit is given by τ
τ = RC = (500) (1.5 × 10⁻⁶) = 0.00075 s
c) The current decay in an RC circuit (called decay because the current in the circuit starts to fall as the capacitor's voltage rises as the capacitor charges) is given by
I = I₀ e⁻ᵏᵗ
where k = (1/τ)
I₀ = Current in the circuit at t = 0 s; I₀ = 0.01232 A
At t = τ = 0.00075 s, kt = (τ/τ) = 1
I = 0.01232 e⁻¹ = 0.0045323 A = 4.532 mA
d) The voltage for a charging capacitor is given by
Vc = Vs (1 - e⁻ᵏᵗ)
where k = (1/τ)
At t = τ = 0.00075 s, Vc = ?, Vs = 6.16 V, kt = 1
Vc = 6.16 (1 - e⁻¹) = 6.16 (0.6321)
Vc = 3.894 V
