1.Consider the build-up of the charge on a capacitor as described in the introdu

Question

1.Consider the build-up of the charge on a capacitor as described in the introducation to this lab. Assume that you start with an initially uncharged capacitor, as in figure 1 and then you close the switch to a battery of Voltage V0. After that time, how many time constants must elapse before the charge reaches 39%, 63%, and 86% of its final value?

2.For the same case you were considering in problem 1, asume that the capacitor is now fully charged. Now the switch is flipped to position 2, and the charge on the capacitor begins to decay away. After that time, how many time constants much elapse before the charge reaches 61%, 37% and 13% of its initial value (the value right before you flipped the switch to position 2)?

Explanation / Answer

1)

Q = Q0*(1-e^-t/T)

for Q = 39 percent of Q0 = 0.39*Q0,

so, 0.39 = (1-e^-t/T)

So, -ln(0.61) = t/T

So, t = 0.5*T <--- half time constant (answer)

Similarly, for 63 percent,

t = -ln(0.37) = T < ----- one time constant

for 86 percent,

t = -ln(0.14) = 2*T <-----2 time constants

2)

for 61 percent ,

t = ln(0.61) = 0.5* T <----half time constant

for 37 percent,

t = 1 time constant

for 86 percent,

t = 2 time constants

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