Confusion concerning a test question - explanation needed

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renzworldc

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confusion

hello.. all plz help solve this problem
Q. Steady state is reached with switch (S) open. Switch S is closed at t=0. At t=∞, the voltage marked V is given by
a.1v
b.2v
c.3V
d.1.5V
the ans is 3V. .. but I dnt know how??
 

Re: confusion

Convert Norton to Thevenin gives 9V with 3k Rth. This is driving 1k + 2k voltage divider. This gives V of 2k resistor, and hence across the capacitor, as 9*2/(4+2) = 3V. At t=0 Vc is 9V when switch is closed Vc will drop to 3 V.

Added after 4 minutes:

Sorry the k should just be ohms
 

confusion

thnks a lot .... but how is the voltage accross the capacitor found to be 9V....Is dat so that after steady state is reached the capacitor voltage equals the source voltage??
 

Re: confusion

When switch is open the circuit is a series RC circuit driven by a 9V voltage source. For this case the capacitor would charge to the applied voltage(9V) starting from zero volts in a exponential fashion. In an exponential curve the steady state value indicates fully charged to the applied voltage. As the applied voltage is 9V, the capacitor voltage is also 9V.
So the answer to your query is yes.
 

confusion

If no other questions exist related to the circuit, some parts are apparently only in it to test your understanding. For t->∞, simply omit the capacitor and replace the switch by a short.
 

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