Two capacitors, two resistors and a battery

[Theelectricchild]

Hey guys I am a bit stuck as to what to do here... we haven't covered much on RC circuits yet and id like to learn how to do this properly.

Two 6.0µF capacitors, two 2.2 kiloOhm resistors, and a 12 V source are connected in SERIES. Starting from the uncharged state how long does it take for the current to drop from its initial value to 1.50 mA?

i KNow i need to use the exponential function, but what must i do with the fact that its in series? Any help would be appreciated thanks!

 

[quantumdude]

What's the equivalent resistance of two resistors in series?

What's the equivalent capacitance of two capacitors in series?

Answer those, and you can reduce the circuit to a simple RC circuit.

 

[M98Ranger]

Hi there,

Thank you for reaching out and sharing your question with us. It's great that you are looking to learn and understand RC circuits better.

To solve this problem, we will need to use the equation for the time constant (τ) of an RC circuit, which is given by τ = RC, where R is the total resistance and C is the total capacitance in the circuit. In this case, the total capacitance is the sum of the two 6.0µF capacitors, which gives us 12.0µF. The total resistance is the sum of the two 2.2 kiloOhm resistors, which gives us 4.4 kiloOhms.

Now, since the circuit is in series, the current flowing through each component will be the same. This means that the initial current is equal to the current at any given time, which we will call I. So, using Ohm's Law, we can write I = V/R, where V is the battery voltage and R is the total resistance. Substituting in the values, we get I = 12 V/4.4 kiloOhms = 2.727 mA.

Now, we can use the equation for the charge on a capacitor (Q) at any given time (t), which is given by Q = Q0e^(-t/τ), where Q0 is the initial charge on the capacitor and τ is the time constant. Since the capacitors are initially uncharged, Q0 = 0. So, we can rearrange the equation to get t = -τ ln(Q/Q0). Substituting in the values, we get t = -4.4 kiloOhms * ln(1.50 mA/2.727 mA) = 3.07 seconds.

Therefore, it will take approximately 3.07 seconds for the current to drop from its initial value to 1.50 mA in this circuit.

I hope this helps you understand how to approach and solve this type of problem. If you have any further questions or need clarification, please don't hesitate to ask. Keep up the good work in your studies!