A 9 V battery is connected in series with a 100 Ω resistor and 12µF capacitor. The capacitor is initially uncharged. What is the capacitive time constant of the circuit? How much charge q(t) is in the capacitor after two time constants?. What is the limiting value of the charge on the capacitor? When the battery is disconnected and the capacitor is discharged, what is the current i(t) in the circuit after two time constants? What is the limiting value of the current? Draw the curves q(t) and i(t) as functions of time and label them appropriately.
The time constant is RC and that the initial capacitor charge is q(0) = CV. That is basic stuff.
Show us what you know, look up and learn the equations, and don't expect us to do all the work here. You will learn the subject better that way.
To find the capacitive time constant of the circuit, you need to know the resistance and capacitance values. In this case, the resistance is 100 Ω and the capacitance is 12µF. The capacitive time constant (τ) can be calculated using the formula τ = RC, where R is the resistance and C is the capacitance.
Using the given values, the capacitive time constant can be calculated as follows:
τ = (100 Ω) * (12µF)
= 1.2 ms
To calculate the charge on the capacitor after two time constants, you need to use the formula q(t) = Q (1 - e^(-t/τ)), where t is the time and Q is the maximum charge that the capacitor can hold. Since the capacitor is initially uncharged, Q can be determined using the formula Q = CV, where C is the capacitance and V is the voltage across the capacitor. In this case, V is the battery voltage, which is 9V.
So, Q = (12µF) * (9V)
= 108µC
Using the formula q(t) = Q (1 - e^(-t/τ)), you can substitute the values to find the charge on the capacitor after two time constants:
q(2τ) = (108µC) (1 - e^(-(2τ)/τ))
= (108µC) (1 - e^-2)
≈ 98.33µC
The limiting value of the charge on the capacitor is the maximum charge it can hold, which is Q. So, the limiting value of the charge is 108µC.
To find the current (i(t)) in the circuit after two time constants when the battery is disconnected and the capacitor is discharged, you can use the formula i(t) = (V / R) * e^-(t/τ), where V is the battery voltage and R is the resistance.
Substituting the values, you get:
i(2τ) = (9V / 100 Ω) * e^(-(2τ)/τ)
= (9V / 100 Ω) * e^-2
≈ 0.590 mA
The limiting value of the current is zero since the capacitor is fully discharged when the battery is disconnected.
To draw the curves q(t) and i(t) as functions of time, you can plot them on a graph. The x-axis represents time (t), and the y-axis represents charge (q) or current (i). Label the curves appropriately, and make sure to include proper units.
The q(t) curve will start at zero and gradually increase towards the limiting value of 108µC. The i(t) curve will start at a maximum value and decay exponentially towards zero. Both curves will have a time constant of 1.2 ms, which represents the rate at which they change.