Physics, asked by Vishal101100, 9 months ago


<body bgcolor=lime><marquee><font color=red>
solve attachment.....​

Attachments:

Answers

Answered by kailashmeena123rm
14

\huge \:  \red   \star \: {  \large \pink {  \underline\mathcal{ANSWER}} }  \:  \red\star \:  \: </p><p></p><p> \:

  • when a capacitor, resistance, battery and key connected in series and key is closed, then
  • charge

v \:  =  \: \: v _ c + 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 helps

Answered by rk4846336
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

Similar questions