Find the initial current in the circuit (at t=0)

[Fatima Hasan]

Homework Statement

In the circuit shown,the initial charge on the capacitor is Q₀ = 200μC . If the switch is closed at t=0 , the capacitor starts discharging through the resistance R=4MΩ . What is the initial current (in μA) (i.e. at t=0) in the circuit?

Homework Equations

I = I₀ e^-t/τ
τ = R_eq *C
Q = CΔV

The Attempt at a Solution

When the switch is closed at t=0 , no current will flow .
I = 0
Is my answer correct ?

 

[haruspex]

The initial current

Initial charge.

When the switch is closed at t=0 , no current will flow .

Why not? What is the voltage?

 

[ehild]

Homework Statement

The initial current on the capacitor is Q₀ = 200μC . If the switch is closed at t=0 , the capacitor starts discharging through the resistance R=4MΩ . What is the initial current (in μA) (i.e. at t=0) in the circuit?

What is the capacitance of that capacitor? And the 200 μC is charge, not current.

The Attempt at a Solution

When the switch is closed at t=0 , no current will flow .
I = 0
Is my answer correct ?

No. The current is zero when the switch is open, but starts to flow just at the instant it is closed. How does the current change with time in case of a discharging capacitor?

 

[Fatima Hasan]

Initial charge.

Why not? What is the voltage?

What is the capacitance of that capacitor? And the 200 μC is charge, not current.

No. The current is zero when the switch is open, but starts to flow just at the instant it is closed. How does the current change with time in case of a discharging capacitor?

τ =R_eq C
= 4*10⁶ * 2*10^-6
τ= 8s
Using KVL :
-RI + V_c = 0
Vc = Q / C
- 4*10⁶ * I + (200 μ)/(2μ) =0
I = 2.5*10^-5 = 25 μA
I = I₀ e^-t/τ
2.5*10^-5 = I₀ e^0/8
I₀ = 2.5*10^-5
= 25 μA

 

[haruspex]

τ =R_eq C
= 4*10⁶ * 2*10^-6
τ= 8s
Using KVL :
-RI + V_c = 0
Vc = Q / C
- 4*10⁶ * I + (200 μ)/(2μ) =0
I = 2.5*10^-5 = 25 μA
I = I₀ e^-t/τ
2.5*10^-5 = I₀ e^0/8
I₀ = 2.5*10^-5
= 25 μA

Right answer, but it is much simpler than that.
You know the charge and capacitance, so what is the potential? What current results when that potential is placed across that resistance?

 

[ehild]

τ =R_eq C
= 4*10⁶ * 2*10^-6
τ= 8s
Using KVL :
-RI + V_c = 0
Vc = Q / C
- 4*10⁶ * I + (200 μ)/(2μ) =0
I = 2.5*10^-5 = 25 μA
I = I₀ e^-t/τ
2.5*10^-5 = I₀ e^0/8
I₀ = 2.5*10^-5
= 25 μA

it is correct now.