Find how the voltage across the capacitor C varies with time t (Fig.) after the shorting of the switch Sw at the moment t = 0.
Text Solution
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Sol. This is an example of R – C circuit in parallel.
The distribution of charge and current in circuit at an instant t is shown in fig.
In loop (1) (ABFGA),
ξ . (I – I 1 ) R – IR = 0
or ξ – IR + I 1 R – IR = 0
or ξ – 2IR + I 1 R = 0
∴ I = 
In loop BDEFB, – 
+ (I – I 1 ) R = 0
or – 
+ 
R = 0
or – 
+ 
R = 0
or – 
+ 
= 0
or 
– 
= 
But I 1 = 
∴ 
– 
= 
or ξ C – 2q = 
or 
= 
After integrating, we get
q = 
(1 – e –2t/RC )
or 
= 
(1 – e –2t/RC )
∴ V = 
(1 – e –2t/RC )
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