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Electronics/mathematics question?
Hi people, can you please help me here, I have to work out
the supply voltage across the capacitor after one time constant as a percentage of the supply voltage.
My r*c is 470x10^-6 * 160=47 and the supply voltage is 9V
I have no idea how to do this, so of someone can guide me it will be appreciate
Thank you
1 Answer
- 8 years agoFavorite Answer
When the capacitor is being charged,then the the formula will be
V(t)=V(1-exp^-t/RC)
Where V is the supply/initial voltage
t is the time duration
RC=time constant
now,in your question t=RC( because it is given after one time constant)
therefore, V(t)=5.689 Volts
and It is easy to find 5.689 is 63.212 % of 9 .
Hence, After one time constant, voltage across capacitor will 63.212% of supply voltage.
Please note,while the capacitor is getting discharged,then the formula will change to
V(t)=V(exp^-t/RC)
Source(s): self
