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Question
Find the equivalent capacitances of the combinations shown in figure between the indicated points.
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Solution
Applying star-delta conversion in the part indicated in the diagram.
The Capacitance of the `C_1` is given by
`C_1^' = (C_2C_3)/(C_1+C_2+C_3)`
`C_2^' = (C_1C_3)/(C_1+C_2+C_3)`
`C_3^' = (C_1C_2)/(C_1+C_2+C_3)`
⇒ `C_1'^ = (3 xx 4)/(1+3+4) = 12/8 "uF"`
⇒ `C_2^' = (1 xx 3)/(1+3+4) = 3/8 "uF"`
⇒ `C_3^' = (1 xx 4)/(1+3+4) = 4/8 "uF"`
Thus, the equivalent circuit can be drawn as :
Therefore, the equivalent capacitance is given by
`C_(eq) = 3/8 + [((3+1/2) xx (3/2+1))/((3+1/2)+(3/2+1))]` = `3/8 + 35/24 = (9+35)/24 = 11/6 "uF"`
`1+1 = 2 "uF"`
(c)
It is a balanced bridge.
Therefore, the capacitor of capacitance 5 μF can be removed.
Cef = `4/3+8/3+4 = 8 "uF"`
The capacitance of the two branches are :
`C_1 = (2 xx 4)/(2+4) = 4/3 uF`
`C_2 = (4 xx 8)/(4+8) = 8/3 uF`
∴ Equivalent capacitance = `4/3 + 8/3 + 4 = 8 "uF"`
It can be observed that the bridges are balanced.
Therefore, the capacitors of capacitance 6 μF between the branches can be removed
The capacitances of the four branches are :
`C_1 = (2 xx 4)/(2+4) = 4/3 uF`
`C_2 = (4 xx 8)/(4+8) = 8/3 uF`
`C_3 = (4 xx 8)/(4+8) = 8/3 uF`
`C_4 = (2 xx 4)/(2+4) = 4/3 uF`
∴ Equivalent capacitance
= `4/3 + 8/3 + 8/3 + 4/3`
= ` 8 "uF"`
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