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The minimum value of effective capacitance that can be obtained by combining 3 capacitors of capacitances 1 pF, 2 pF and 4 pF is

Answer»

`(4)/(7)PF`
1 pF
`(7)/(4)pF`
2 pF

Solution :The minimum VALUE of EFFECTIVE CAPACITANCE is get in series combination. Now, the equivalent capacitance,
`(1)/(C_(eq))=(1)/(1)+(1)/(2)+(1)/(4)`
`(1)/(C_(eq))=(4+2+1)/(4)`
`(1)/(C_(eq))=(7)/(4)`
`C_(eq)=(4)/(7)pF`


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