What is the Capacitance of the Unknown Capacitor in the Circuit?

[xswtxoj]

Homework Statement

a series combinations is 11.6k ohms resistor and an unknown capacitor is connected to the a 180 V battery. One second after the circuit is completed, the voltage across the capacitor is 16.8v. determine the capacitance of the capacitor?

Homework Equations

V=18
C?
1 sec later
V=16.8v

The Attempt at a Solution

c= q/v
i=v/r

 

[swuster]

Derive the differential equation for the current/voltage across the capacitor when it's charging, using Kirchoff's Voltage Law, and solve.

Or if you're allowed to, just look up the solution of this equation, which should be in the form $$V_c (t) = V_o (1 - e^{-t/\tau})$$ where $$\tau$$ is the time constant $$RC$$. You can then plug in your known values of $$V_c$$ to solve for the capacitance.

 

[nlightner]

Thank you for reaching out for help with series combinations. From the information provided, we can use the formula V=IR to determine the current flowing through the circuit. We know that the voltage across the resistor is 11.6k ohms and the battery is 180V, so we can solve for I by dividing 180V by 11.6k ohms. This gives us a current of approximately 0.0155A.

Next, we can use the formula V=Q/C to determine the capacitance of the unknown capacitor. We know that the voltage across the capacitor is 16.8V, so we can rearrange the formula to solve for C. This gives us C=Q/V, where Q is the charge stored on the capacitor. We can calculate Q by multiplying the current (0.0155A) by the time (1 second). This gives us a charge of approximately 0.0155C.

Plugging in these values, we get C=0.0155C/16.8V, which simplifies to approximately 0.0009F or 900μF.

I hope this helps. Let me know if you have any further questions or if you would like me to explain anything in more detail. Good luck with your homework!