Why do you need to start recording in Part (A) simultaneously with the DC power

Question

Why do you need to start recording in Part (A) simultaneously with the DC power supply? Mow much voltage (in terms of the power source voltage v_b) will the capacitor have when it has started at zero volts potential difference, it is connected to the power supply and resistor and one characteristic time Ti as past (i.e. t = tau)? Keep a copy - you will need this during vour lab and in vour report. Mow much voltage (m terms of the initial charge on the capacitor) will the capacitor have when it is connected to a resitor and one characteristic time has past (i.e. t = tau)? Keep a copy - you will need this during your lab and in your report.

Explanation / Answer

A.) In part, that is while charging the capacitor, what do you think is the time it takes to charge it completely?

Look at the equation VC = Vb ( 1 - e-t/RC )

theoretically VC will be equal to Vb only if t = infinity!

It means that it actually takes infinite time to charge the capacitor completely!

But actually we see that most of the charging (more than 90 % of the charging) happens with in 2-3 time constants. This period of time is very less. In fact, given the resistance is high, it vcan be just a fraction of second! So, calculating the time constant is very crucial in this experiment. Even a small error or delay in pressing the start or stop button can lead to a huge error. Because, the entire time taken before we stop recording (to reach 1.9 V) itself is very less. We cannot afford an error of, say, even 1 second in this aspect of measuring time. That is why we need to start recording simultaneously to avoid error, noting the fact that charging also start instantaneously as soon as you connect the DC supply.

B.) The equation for charging of a capacitor is

VC = Vb ( 1 - e-t/RC  )

put t=0 , we get Vc = 0 , that is initial potential difference across the capacitor is zero as given.

at t= RC, that is at one time constant,

the equation becomes , VC = Vb ( 1 - e-RC/RC ) = Vb ( 1 - e-1 ) = Vb ( 1 - 1/e ) = Vb ( 1 - 0.367879441)

VC = Vb ( 0.632120558) = 0.6321 Vb   i.e 63.21 % of Vb

C.) The capacitor is connected to a resistor and note the words ' in terms of the initial charge' which clearly says that the capacitor was initially charged. So, this is a case of discharging of a charged capacitor.

VC = Vo e-t/RC

Where Vo is the initial voltage of the fully charged capacitor, C = Qo /Vo  where Qo is the initial charge

Therefore, VC = (Qo /C ) x e-t/RC

After, one time constant t =RC, putting this in the above equation,

VC = (Qo /C ) x e-t/RC = (Qo /C ) x e-RC/RC = (Qo /C ) x e-1 = (Qo /C) x (1/e) = 0.367879441 Qo /C

VC ~ 0.367879441 Qo /C

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