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In the circuit shown in figure, find the...

In the circuit shown in figure, find the steady state charges on both the capacitors.

Text Solution

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In the steady state, there will be no current flowing through the two capacitors, Therefore, current in lower circuit ABCD is

`i_(1) = (20)/(6+4) = 2A`,
and current in upper circuit EFGH is
`i_(2) = (10)/(2+3) = 2A`
Obviously, charges on both capacitors will be same, say q each.
Applying Kirchhoff's law to the loop ADEHAlt we get
`4i_(1) - (q)/(6xx10^(-6)) - 3i_(2) - (q)/(3xx10^(-6)) = 0`
or `4i_(1) - 3i_(2) = (q)/(6xx10^(-6)) + (q)/(3xx10^(-6))`
or `4xx2-3xx2 = (3q)/(6xx10^(-6)) = (q xx 10^(-6))/(2)`
`q xx 10^(6) = 2+2 = 4`
`q = (4)/(10^(6)) c = 4xx10^(-6)c = 4 muC`
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