210k views
3 votes
. 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?

2 Answers

4 votes

Answer:

Step-by-step explanation:

Given an RC circuit to analyze

R=500Ω

C=1.50-μF uncharged

Emf(V)=6.16V

Series connection

a. Initial current, since the capacitor is initially uncharged then, the voltage appears at the resistor

Using ohms law

V=iR

Then, i=V/R

i=6.16/500

i=0.01232 Amps

i=12.32 mA.

b. The time constant is given as

τ=RC

τ=500×1.5×10^-6

τ=0.00075second

τ=0.75 ms

c. What is current after 1 time constant

Current in a series RC circuit is given as

Time after I time constant is

t=1 ×τ

t= τ

i=V/R exp(-t/RC)

Where RC= τ

i=V/Rexp(-t/ τ)

i=6.16/500exp(-1), since t= τ

i=0.004532A

I=4.532mA

d. Voltage after one time constant

Voltage of a series RC circuit(charging) is given as

Again, t= τ

V=Vo(1 - exp(-t/ τ)),

V=6.16(1-exp(-1))

V=6.16(1-0.3679)

V=6.16×0.632

V=3.89Volts

V=3.89V

V=

User GlyphGryph
by
5.6k points
5 votes

Answer:

initial current I₀ = 0.0123 A

RC time constant τ = 0.00075 sec

current after one time constant I = 0.00452 A

voltage on the capacitor after one time constant V = 3.89 V

Step-by-step explanation:

Given that,

Voltage = 6.16 V

Resistance = 500 Ω

Capacitance = 1.5 µF

(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

User Moslem Shahsavan
by
5.3k points