# Voltage drop across a capacitor

• amolv06
In summary, the conversation discusses a test question involving finding the current leaving a battery using Kirchoff's laws and the equations V=IR and C=Q/V. The issue is that the voltage drop across the capacitor is unknown and the attempt at solving it involves using a loop equation. The solution may also depend on whether the problem is in steady-state or if there are initial conditions to consider.
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?

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:

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

#### Attachments

• schem.JPG
5.7 KB · Views: 1,321
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:

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

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

## 1. What is voltage drop across a capacitor?

Voltage drop across a capacitor refers to the decrease in voltage that occurs across the capacitor as it charges and discharges. This is due to the capacitor's ability to store and release electrical energy, which affects the voltage level in a circuit.

## 2. How is voltage drop across a capacitor calculated?

The voltage drop across a capacitor can be calculated using the formula V = Q/C, where V is the voltage drop, Q is the charge on the capacitor, and C is the capacitance of the capacitor. This formula is based on the relationship between voltage, charge, and capacitance in a capacitor.

## 3. What factors affect the voltage drop across a capacitor?

The voltage drop across a capacitor can be affected by various factors, including the capacitance of the capacitor, the amount of charge stored on the capacitor, and the frequency of the AC signal passing through the circuit. The type and material of the capacitor can also impact the voltage drop.

## 4. Why is voltage drop across a capacitor important to consider?

The voltage drop across a capacitor is important to consider because it can affect the performance of a circuit. If the voltage drop is too high, it can cause the circuit to malfunction or not function at all. It is also important for determining the energy storage and release capabilities of a capacitor.

## 5. How can the voltage drop across a capacitor be reduced?

The voltage drop across a capacitor can be reduced by increasing the capacitance of the capacitor or reducing the amount of charge stored on it. Using a capacitor with a lower frequency rating or using multiple capacitors in parallel can also help reduce the voltage drop. Additionally, using higher quality capacitors with lower internal resistance can also minimize the voltage drop.

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