A capacitor C 1 is charged to a potential difference V 1 . Its terminals are the
ID: 1323930 • Letter: A
Question
A capacitor C1 is charged to a potential difference V1. Its terminals are then connected though a resistence R to the terminals of an uncharged capacitance C2. Write the differential equation for the circuit, and find V across the first capacitor at any time t after it is connected to the second capacitor.
Hint: If you let q1 be the variable charge on C1 at any time t, what is the charge on C2 at any time t (in terms of q1) ?
A capacitor C1 is charged to a potential difference V1. Its terminals are then connected though a resistence R to the terminals of an uncharged capacitance C2. Write the differential equation for the circuit, and find V across the first capacitor at any time t after it is connected to the second capacitor. Hint: If you let q1 be the variable charge on C1 at any time t, what is the charge on C2 at any time t (in terms of q1) ? Ans: left [ C_{1}+C_{2}e^{frac{ - t}{RC_{1}C_{2}/(C_{1}+C_{2})}} right ] frac{V{_{1}}^{}}{(C_{1}+C_{2})}Explanation / Answer
When we picture the circuit, it will look like the capacitors in seris.
So the net Ceq of the circuit will be Ceq = C1C2/ (C1+C2)
V = VR + VC = IR + Qtot/C where I = V1/R
So, V = V1 + Qtot/C eq (1)
As we know that, when the charging is in progress, the current decreases and Q increases and Q at any time after it is connected will be given by
Qtot = CeqV1 = CeqV1 (e -t/RC1C2(C1+C2 ) = C1C2/ (C1+C2)V1 (e -t/RC1C2(C1+C2) )
Putting this value in the eq (1), we get
V = V1 + C1C2/ (C1+C2)V1 (e -t/RC1C2(C1+C2 )
V = V1/(C1+C2) [ C1 + C2(e -t/RC1C2(C1+C2 ) ]
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