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?

Respuesta :

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.

A )The initial current in the circuit ( I₁ ) = 0.0123 A

B) The RC time constant is = 7.5 * 10⁻⁴ sec  

C) The current after  one time constant = 0.00452 A

D) The voltage on the capacitor after the one time constant = 3.89 V

Given data :

Resistance  ( R ) = 500 -Ω

Capacitance of the capacitor ( C ) = 1.50-μF

Voltage source = 6.16-V

A) Calculate the Initial current  found in the circuit

I₁ = voltage / resistance of resistor

  = 6.16 / 500 = 0.0123 A

B) Calculate for the value of RC time constant ( i.e. time to fully charge capacitor)

RC time constant ( i )  = R * C

                             = 500 * 1.50 * 10⁻⁶ = 7.5 * 10⁻⁴ sec.

C) Determine current after one time constant

I = [tex]I_{1} e^{(-τ/RC)}[/tex] --- ( 1 )         given that i = Rc

I = 0.0123 * e^(-1)

∴ I ( current after one time constant )  =  0.00452 A

D) Voltage on capacitor after one time constant

V = V₀(1 - e^(-τ/RC))

   = 6.16 ( 1 - e^(-1) )

   = 6.16 ( 1 - 0.3679 )

   = 3.89 V

Hence we can conclude that The initial current in the circuit ( I₁ ) = 0.0123 A The RC time constant is = 7.5 * 10⁻⁴ sec , The current after  one time constant = 0.00452 A and The voltage on the capacitor after the one time constant = 3.89 V .

Learn more : https://brainly.com/question/22428024

Ver imagen batolisis
Q&A Education