RC-RL Circuit: Solving for Capacitance Value

[paulmdrdo]

Homework Statement

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2. Homework Equations

3. The Attempt at a Solution

In got stuck. I'm not sure as to what Time constant should I use to solve for the capacitance value. In the original expression for the current, both terms are exponential function. I do not know whether to use the time constant associated with the first term or the second term.

 

[Delta2]

The voltage of the source is given by $V_s(t)=36e^{-t}u(t)$ ?? What function is $u(t)$?

 

[paulmdrdo]

The voltage of the source is given by $V_s(t)=36e^{-t}u(t)$ ?? What function is $u(t)$?

That's a unit step function.

 

[Delta2]

Well ok, I am not 100% sure but I believe the second exponential contains info about the capacitance. More specifically I think that $\frac{1}{(R_1+R_2)C}=0.6$

 

[paulmdrdo]

Well ok, I am not 100% sure but I believe the second exponential contains info about the capacitance. More specifically I think that $\frac{1}{(R_1+R_2)C}=0.6$

Thank you. That's what I thought too. But It would be nice if someone would explain as to why that is the case.

 

[Delta2]

Thank you. That's what I thought too. But It would be nice if someone would explain as to why that is the case.

.

By setting up and solving the differential equation for this circuit and using that $V_s(t)=36e^{-t}$ I found that the current is $$I(t)=ae^{-t}+be^{-\frac{t}{(R_1+R_2)C}}(1)$$ for some constants a and b and by comparing this with the equation you give for current, I decided my conclusion.

 

[NascentOxygen]

If there was zero initial charge on the capacitor, then the current magnitude at t=0 would be 36 ÷ 14 amps. But we are told the current is a spike or pulse of magnitude 9 A peak at t=0. So we conclude there must be some initial charge on the capacitor in order to give a total loop voltage at t=0 of 9 ×14 volts.

If you set up and solve the 1st order D.E. this should be confirmed.