Finding voltage and current expressions (capacitance)

[appleraja]

Homework Statement

The charge on a 2-microfarad capacitor is given by q(t)=10^-6 * sin(10^5*t) C find the dexpression for the voltage and current?

Homework Equations

The Attempt at a Solution

the answer is v(t)=0.5sin(10^5t) , i(t)=0.1cos(10^5t) i don't get how they found the answer

 

[grzz]

Note that q = Cv and i = $\frac{dq}{dt}$.

 

[appleraja]

what I am confused is when they say, charge is microfarad does that mean q=2x10^-6 or is the capacitor 2 microfarad meaning c=2x10^-6

 

[gneill]

what I am confused is when they say, charge is microfarad does that mean q=2x10^-6 or is the capacitor 2 microfarad meaning c=2x10^-6

Capacitance is given in farads (microfarads in this case), while charge is given in coulombs (C).

 

[rezeew]

As a scientist, it is important to understand the concepts and equations that govern electrical circuits and components. In this case, the capacitor is a key component and its behavior is described by the relationship between charge, voltage, and current.

The given charge expression, q(t)=10^-6 * sin(10^5*t) C, represents the charge on the capacitor as a function of time. To find the voltage and current expressions, we can use the equations q=CV and i=dq/dt, where C is the capacitance and i is the current.

Substituting the given charge expression into the first equation, we get q(t)=CV, which can be rearranged to V=q(t)/C. Since the capacitance is given as 2 microfarads, or 2*10^-6 farads, the voltage expression becomes V(t)=q(t)/(2*10^-6).

Next, we can use the second equation, i=dq/dt, to find the current expression. Taking the derivative of the given charge expression, we get dq/dt=10^5*cos(10^5*t). Substituting this into the equation for current, i=dq/dt, we get i(t)=10^5*cos(10^5*t). However, this expression is in terms of microamps (10^-6 amps), so we need to multiply it by 10^-6 to get the final current expression: i(t)=10^-6 * 10^5*cos(10^5*t) = 0.1*cos(10^5*t).

By understanding the concepts and equations that govern capacitor behavior, we can easily find the voltage and current expressions for this circuit. It is important to always check the units and make sure they are consistent in our calculations.