An RC circuit contains a 3.9 kΩ resistor and a 5 µF capacitor. When the switch is first closed in the circuit, a maximum current of 1.5 mA flows in the circuit. Using Equation (1), find the time at which the current flow will have fallen to 1.0 mA.
Hmmm.
I=V/R e^-t/rc is that you equation?
1.0=3.5e^t/rc
take the ln of each side
ln1.0-ln3.5=t/RC
solve for time t.
1.5/e^x = 1.0
e^x = 1.5
X = 0.405 = t/RC.
RC = 3.9*5 = 19.5 milliseconds.
t/19.5 = 0.405
t = 7.91ms.
To find the time at which the current flow in the RC circuit will have fallen to 1.0 mA, we can use the equation:
I(t) = I(0) * e^(-t/RC) ........... (1)
Where:
- I(t) is the current at time t.
- I(0) is the initial current when the switch is closed (t = 0).
- e is the base of the natural logarithm (approximately 2.71828).
- t is the time in seconds.
- R is the resistance in ohms.
- C is the capacitance in farads.
In this case, we are given:
- I(0) = 1.5 mA (convert to Amperes: 1.5 mA = 1.5 * 10^-3 A)
- R = 3.9 kΩ (convert to ohms: 3.9 kΩ = 3.9 * 10^3 Ω)
- C = 5 µF (convert to farads: 5 µF = 5 * 10^-6 F)
Substituting these values into Equation (1), we can rearrange the equation to solve for t:
t = -RC * ln(I(t) / I(0))
Now, let's calculate the time t at which the current will have fallen to 1.0 mA (convert to Amperes: 1.0 mA = 1.0 * 10^-3 A):
t = - (3.9 * 10^3 Ω) * (5 * 10^-6 F) * ln(1.0 * 10^-3 A / (1.5 * 10^-3 A))
Calculating this expression will give you the time at which the current flow will have fallen to 1.0 mA.