AC Circuit Analysis - Capacitor

[Gorz]

Homework Statement

Calculate the magnitude and sign of i₁₂ which will flow through a capacitor of 10microfarads when:

v₁₂ = +50V, and decreasing at a rate of 10⁴V/s

Homework Equations

Vab=Q/C

The Attempt at a Solution

No idea where to even begin with this question I am affraid. :(

 

[Gorz]

Solved, please close.

 

[nlightner]

I would first start by identifying the given information and what is being asked for. The problem states that we have a capacitor with a capacitance of 10 microfarads and we are given the voltage across the capacitor, v12, which is +50V and decreasing at a rate of 104V/s. The problem is asking us to calculate the magnitude and sign of the current, i12, that will flow through the capacitor.

To solve this problem, we can use the equation Vab=Q/C, where Vab is the voltage across the capacitor, Q is the charge on the capacitor, and C is the capacitance. We can rearrange this equation to solve for Q, which will give us the charge on the capacitor.

Q = Vab * C

Substituting in the given values, we get:

Q = (+50V) * (10 microfarads)

= 500 microcoulombs

Now, we need to find the current, which is the rate of change of charge with respect to time. In other words, it is the change in charge over a certain period of time. We can use the equation for current, i = ΔQ/Δt, where ΔQ is the change in charge and Δt is the change in time.

Since we are given the rate of change of voltage, we can use this to find the change in time. We know that the voltage is decreasing at a rate of 104V/s, so we can say that the change in time is 1 second, as the voltage is decreasing by 104V in 1 second.

Therefore, the current can be calculated as:

i12 = ΔQ/Δt

= (500 microcoulombs) / (1 second)

= 500 microamps

Since the voltage is decreasing, the current will be flowing in the opposite direction of the voltage, so the sign of the current will be negative. Therefore, the magnitude and sign of i12 is 500 microamps in the opposite direction of the voltage, or -500 microamps.