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Question
Find the capacitance of the combination shown in figure between A and B.
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Solution
Capacitors 5 and 1 are in series.
Their equivalent capacitance, `C_(eq) = (C_1C_5)/(C_1+C_5)` =`(2 xx 2)/(2+2)` = `1 "uF"`
`therefore` `C_(eq) = 1`
Now, this capacitor system is parallel to capacitor 6. Thus, the equivalent capacitance becomes 1 + 1 = 2 μF
The above capacitor system is in series with capacitor 2. Thus, the equivalent capacitance become `(2 xx 2)/(2+2) = 1 "uF"`
The above capacitor system is in parallel with capacitor 7. Thus, the equivalent capacitance becomes 1 + 1 = 2 μF
The above capacitor system is in series with capacitor 3. Thus, the equivalent capacitance becomes `(2 xx 2)/(2+2) = 1 "uF"`
The above capacitor system is in parallel with capacitor 8. Thus, the equivalent capacitance becomes
1 + 1 = 2 μF
The above capacitor system is in series with capacitor 4. Thus, the equivalent capacitance becomes `(2 xx 2)/(2+2) = 1 "uF"`
Hence, the equivalent capacitance between points A and B of the given capacitor system is 1 μF.
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