[SOLVED] charge sharing in capacitors

[srimatresri]

can you help me in solving this question
after closing the switch what is the voltage at vx at t=0+
vc is the initial voltage of the capacitor

 

[Easyrider83]

In ideal condition you will have a short circuit and impossibly high current. In a real - depends of ESR of capacitors and supply resistance.

 

[srimatresri]

what will be voltage if its ideal condition?

 

[Easyrider83]

what will be voltage if its ideal condition?

0 Volts. In ideal condition uncharged capacitor will be like a short circuit.
Except C3. 0.8V will stays. :-D

 

[srimatresri]

the voltage at vx is initially 0.8 v before the switch is closed

 

[Easyrider83]

the voltage at vx is initially 0.8 v before the switch is closed

Sorry, it will 0.8V like you said. I didn't understand what vx means first time.

 

[timof]

0.8v + 4.2v * 2/17 ~ 1.294 V ?

 

[srimatresri]

0.8v + 4.2v * 2/17 ~ 1.294 V ?

what i think is , infinite current flows in a small amount of time and the vx will be 1.294V immediately

 

[srimatresri]

0.8v + 4.2v * 2/17 ~ 1.294 V ?

but sir,
if we take from the perspective of time response, t=0+ represents the transient response , and the 1.294 v is the steady state value. so if v(t=0+ )=1.294 V instead of 0.8 V, it has bypassed transient and reached steady state?
i am really confused.

 

[saro_k_82]

There are no resistors in the circuit to delay the steady state. So as it is, v(t=0+) should give 1.294V. Practically there will be a small interval between 0.8V and 1.294V due to the time constants determined by capacitors and the ESR of the caps and wiring resistance.