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Answered by
14
- when a capacitor, resistance, battery and key connected in series and key is closed, then
- charge
v = q/c + ir = q/c + dq/dt × r
q = cv[1 - e ^( -t/rc )] = qo [1 - e^(-t/rc )]
at t = tow = rc = time constant
q = qo [1 - e^-1 ] = 0.632 qo
so, in charging, charge incrreases to 63.2 % of charge in the time equal to t
- current
I = dq/dt = io e ^ ( e^(-t/rc ))
- potential
v = vo (1-e^(-t/rc))
hope it helps
Answered by
7
Explanation:
when a capacitor, resistance, battery and key connected in series and key is closed, then
charge
v \: = \: \: v _ c + v _ rv=vc +v r
v = q/c + ir = q/c + dq/dt × r
q = cv[1 - e ^( -t/rc )] = qo [1 - e^(-t/rc )]
at t = tow = rc = time constant
q = qo [1 - e^-1 ] = 0.632 qo
so, in charging, charge incrreases to 63.2 % of charge in the time equal to t
current
I = dq/dt = io e ^ ( e^(-t/rc ))
potential
v = vo (1-e^(-t/rc))
hope it helpful for you
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