Question

A 500-Ω resistor, an uncharged 1.50-μF capacitor, and a 6.16-V emf are connected in series. (a) What is the initial current? (b) What is the RC time constant? (c) What is the current after one time constant? (d) What is the voltage on the capacitor after one time constant?

Answer 1

Given Information:

Voltage = 6.16 V

Resistance = 500 Ω

Capacitance = 1.5 µF  

Required Information:

Initial current = I = ?

Time constant = τ = ?

Current after 1 time constant = ?

Voltage after 1 time constant = ?

Answer:

I₀ = 0.0123 A

τ = 0.00075 sec

I = 0.00452 A

V = 3.89 V

Explanation:

(a) What is the initial current?

The initial current can be found using

I₀ = Voltage/Resistance

I₀ = 6.16/500

I₀ = 0.0123 A

(b) What is the RC time constant?

The time constant τ provides the information about how long it will take to charge the capacitor.

τ = R*C

τ = 500*1.5x10⁻⁶

τ = 0.00075 sec

(c) What is the current after one time constant?

I = I₀e^(-τ/RC)

I = 0.0123*e^(-1)   (0.00075/0.00075 = 1)

I = 0.00452 A

(d) What is the voltage on the capacitor after one time constant?

V = V₀(1 - e^(-τ/RC))

Where V₀ is the initial voltage 6.16 V

V = 6.16(1 - e^(-1))

V = 6.16*0.63212

V = 3.89 V

That means the capacitor will charge up to 3.89 V in one time constant.

Answer 2

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