Question

what is the capacitance of a condenser c if the resultant capacitance between points a and b is 12uF​

Answer 1

Answer:

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The equivalent capacitance for RS is CRS=4+6+126×12=4+4=8μF

Now, CPS=1+81×8=8/9μF and CPQ=8+48×4=8/3μF

Now CPQ and CRS are in parallel and then it is series with C.

Thus, CAB=C+32/9C(32/9)=1

or C+32/9=32C/9

or 9C+32=32C or 23C=32 

so, C=32/23=1.4μF

Answer 2

According to the circuit,

case-1: capacitor a and b are connected in sereies.

case-2: capacitor c and d are connected in series.

case-3: case -1 and 2 are connected in parallel with AB as well as with the C capacitor.

Now,

For case-1:

$\small{ \frac{1}{C_{s1}} = \frac{1}{a} + \frac{1}{b} = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3} \mu \: F }\\ \\ \therefore \: C_{s1} = 3 \mu \: F \:$

For case-2:

$\small{ \frac{1}{C_{s2}} = \frac{1}{a} + \frac{1}{b} = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3} \mu \: F }\\ \\ \therefore \: C_{s2} = 3 \mu \: F \:$

For case-3:

$C_{eq} = C_{s1} + C_{s2} + C \\ = 3 + 3 + x \\ = 6 + x \: \mu \: F \: \\ \\ C_{eq} = 6 + x \: \mu \: F \:$

But given that,

$C_{eq} = 12 \: \mu \: F \:$

So, we can write that,

$6 + x = 12 \\ \\ = > x = 12 - 6 \\ \\ = > x = 6 \: \mu \: F \:$

So capacitance of capacitor C is = 6 $\mu\:F$