Voltage drop across a capacitor

[amolv06]

Homework Statement

Below, I have attached a schematic that was on a test question for me. (NOTE: R4 = 4*r) We were supposed to find the current leaving the battery in terms of R, C, and Vb. The thing was, I had no idea what the voltage drop across the capacitor would be since there is no q given. How would one solve this?

Homework Equations

Kirchoff's laws
C=Q/V => V=Q/C
V=IR

The Attempt at a Solution

It's a bit hard showing since I would have a hard time drawing in all the loops and such, but I had 3 loops. I just didn't know what the equation for the one going through the top right would be since I'm not given q (or t).

I tried the following for the loop in question:

$$V_{b} - R(I_{2}+I_{3}) - IR_{3} - \frac{q}{c} - R(I_{3} + I_{1})=0$$

 

[berkeman]

Did the problem state that it was in steady-state? If so, the capacitor is an open circuit. If you were given some initial conditions different from the steady state condition, were you asked to calculate the response over time as the circuit moved to the final steady state solution? If so, then you need to solve the differential equation that you get when you put in the current-voltage relationship for a capacitor:

$$i(t) = C\frac {dv(t)}{dt}$$

 

[Moetasim]

I would approach this problem by first understanding the basic principles behind capacitors and how they behave in a circuit. A capacitor is a device that stores electric charge and has the ability to block direct current (DC) while allowing alternating current (AC) to pass through.

In this circuit, the capacitor is connected in parallel with the resistor R3, which means the voltage drop across the capacitor will be the same as the voltage drop across R3. This is because in a parallel circuit, the voltage is the same across all branches.

Using Kirchoff's laws, we can write the following equation for the loop that includes the capacitor:

Vb - R(I2+I3) - IR3 - VC = 0

Where VC is the voltage drop across the capacitor. We can also use Ohm's law to write:

VC = IC * R3

Substituting this into the previous equation, we get:

Vb - R(I2+I3) - IR3 - IC * R3 = 0

Now, we need to find an expression for the current IC in terms of the given parameters. We know that the current through a capacitor is given by:

IC = C * dV/dt

Where C is the capacitance and dV/dt is the rate of change of voltage with respect to time. In this circuit, the voltage across the capacitor is changing as the capacitor charges, so we can write:

dV/dt = (Vb - VC)/R4

Substituting this into the previous equation and rearranging, we get:

IC = (C * Vb - C * VC)/R4

Now, we can substitute this expression for IC into our original equation and solve for VC:

Vb - R(I2+I3) - IR3 - (C * Vb - C * VC)/R4 = 0

Solving for VC, we get:

VC = (Vb * R4 - R * (I2+I3) - IR3)/ (R4 + RC)

Now, we can use this expression for VC to solve for the current leaving the battery, which is given by:

I1 = (Vb - VC)/R

Substituting the expression for VC, we get:

I1 = (Vb - (Vb * R4 - R * (I2+I3) - IR3)/ (R