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# A fixed capacitor of capacitance 10^-4 microfarad is connected in series with a variable capacitor, the capacitance of which may be varried from zero to 10^-4 microfarad in steps of 10^-6 These two in series are connected in parallel with a third capacitor of capacitance 5x10^-4 microfarad. Calculate .1) the maximum capacitance of the whole combination, 2) the smallest change in the capacitance which can be produced by the arrangement.

Vikas TU
14149 Points
one year ago
When first capacitor and second capacitor are connected in series, then the capacitance will be maximum when the value of c2 is taken as 0.0001.
so when cap is in series Ceq = (c1*c2)/(c1+c2)
and when it is in parallel Ceq= c1+c2
so maximum capacitance would be 5.5* 10-4 microfarad.
The smallest change in capacitance can be seen by arrangement , is when c2 is changed from 0 microfarad to 10-6 microfarad.
Hope this wil clear .