Find the equivalent capacitance between A and B in the circuit shown in the Figure. C1 = C4 = 1 μF; C2 = C3 = 2 μF.
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when you arrange it in proper way with help of using short circuit , you will get circuit as shown in figure.
C2 and C3 are in series combination.
so, C' = C2.C3/(C2 + C3)
C' = 2 × 2/(2 + 2) = 1 μF
now, C' , C1 and C4 are in parallel combination.
so, Ceq = C' + C1 + C4
= 1μF + 1μF + 1μF
= 3μF
hence, equivalent capacitance = 3μF
C2 and C3 are in series combination.
so, C' = C2.C3/(C2 + C3)
C' = 2 × 2/(2 + 2) = 1 μF
now, C' , C1 and C4 are in parallel combination.
so, Ceq = C' + C1 + C4
= 1μF + 1μF + 1μF
= 3μF
hence, equivalent capacitance = 3μF
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