In the circuit below, at time t = 0, the switch is closed, causing the capacitors to charge. The voltage across the battery is 12 V

Question

In the circuit below, at time t = 0, the switch is closed, causing the capacitors to charge. The voltage across the battery is 12 V

The circuit:
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A)Calculate the time constant for the RC circuit

B)Calculate the time required for the voltage across the capacitor to reach 6 V

C)What will be the net charge stored in the capacitor after 10 nanoseconds?

What I did:

A) Time constant= [(C1+C2)*(R1R2)]/(R1+R2)
= [(3uF+6uF)*(4k-ohm*2k-ohm)]/(4k-ohm+2k-ohm)
= 12 ms= 0.012 sec?

TC= time constant
B) q= q0[1-e^(-t/TC)]
t= 8.316 ms= 0.008316 sec?

C) I don't know what to do and can't find an equation that I think would work

Answer 1

A) Your answer is correct - the time constant of the circuit is indeed 12ms.

B) The time required to reach the final voltage by a capacitor is roughly 4*TC. In this case,the capacitors would charge up to full battery voltage(6V)in 48 ms.

C) t=10 nano-seconds is such a short time compared to the time constant of the circuit that the capacitors shall hardly be charged. So the net charge stored would be zero coulombs at the end of 10 ns.
You can also calculate the exact amount of q using the formula q=q0[1-e^(-t/TC)]; just plug in q0 = Ceq*V = 9muF*6 and 10ns for t in the equation to get that small value of q.

Answer 2

Thanks for your response, but I have a few questions.

For part B, why is full battery voltage 6V, and not 12V?
For part C, my question is once again about using 6V in the calculation for finding q0, as well as units. Using 6V, qo= 54 uC? For the time constant, I plugged in 12ms, so I changed 10 nanoseconds to milliseconds and plugged the ms value in. Overall, I got 4.5E-5 uC

Answer 3

My mistake - I wrongly considered the battery voltage to be 6V instead of 12V.

B) The capacitor voltage will reach 6V (half of its final voltage) in 0.693*TC time. You can also use the equation:
V = V0{1-e^(-t/RC)} where V=6 and V0=12. You'd get t= 8.316 ms

C)Find q0 by using 12V instead of 6V => q0= 108uC and use the same equation.

Answer 4

To address your questions:

A) To calculate the time constant for the RC circuit, you can use the formula: time constant (TC) = (C1 + C2) * (R1 || R2), where || represents the parallel combination of resistors. In this case, we have C1 = 3uF, C2 = 6uF, R1 = 4kΩ, and R2 = 2kΩ. Plugging these values into the formula:

TC = (3uF + 6uF) * (4kΩ || 2kΩ)
= 9uF * (8kΩ / 6kΩ)
= 12ms

So, your calculation is correct. The time constant for the RC circuit is 12ms.

B) To calculate the time required for the voltage across the capacitor to reach 6V, you can use the formula: t = -TC * ln(1 - V/V0), where t is the time required, TC is the time constant, V is the desired voltage (6V), and V0 is the initial voltage (12V in this case). Plugging in the values:

t = -12ms * ln(1 - 6V/12V)
= -12ms * ln(1/2)
= -12ms * ln(0.5)
≈ 8.316ms

So, the time required for the voltage across the capacitor to reach 6V is approximately 8.316ms.

C) To determine the net charge stored in the capacitor after 10 nanoseconds, you can use the formula: q = C * V, where q is the charge, C is the capacitance (which is 3uF), and V is the voltage across the capacitor. Plugging in the values:

q = 3uF * 12V
= 36uC (microCoulombs)

So, the net charge stored in the capacitor after 10 nanoseconds is 36 microCoulombs.

Note: When referring to equations, it's important to use the appropriate formulas for the given scenario.