Capacitance
Q.
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In the figure, C1 = C5 =8.8 μF and C2 = C3 =C4 = 4.7μF. The applied potential is Vab = 250V.
(a) What is the equivalent capacitance of the network between points a and b?
(b) Calculate the charge on each capacitor and the potential difference across eachcapacitor.
ANS.(a)C3 XX in series with X4 :
the XXXXXXXXXX XXXXXXXXXXX XXX = XXC4/( XX + XX )
= X.X*4.7 /( 4.7 + X.X) = 2.35XX
CX XX in XXXXXX with X34 :
XXX equivalent capacitance CXXX = CX + CXX
=X.7μF +X.XXXX = 7.05μF
Now, C1,CXXX XXX CX are in series :
XXX XXXXXXXXXX Capacitance C=X1*C234*C5/(X1*X234 +X1*X5 + CXXX*XX)
=8.8μF*X.XXXX*X.8μF / (X.8μ*7.XXXX + F8.XXX*8.XXX+7.XXXX*8.XXX)
=2.XX μF
Thus the XXXXXXXXXX XXXXXXXXXXX XX the XXXXXXX XXXXXXX points a XXX b is X.XX μF
(b)The applied potential is Xab = 250V
The XXXXXX on the XXXXXXX X = C*Xab= 2.71μF*XXXX = 677.XX*10-6 X
XX the XXXXXXX XX CX,XXXXXXX X5 XXX equal and = Q =677.XX*XX-X X
The potential difference XXXXXX C1,XX = 677.29*10-XC/8.8μF = XX.97V
The XXXXXXXXX XXXXXXXXXX XXXXXX XXXX,X234= 677.29*XX-XC/X.05μF
= 96.06V
The XXXXXXXXX XXXXXXXXXX across XX,XX= 677.XX*XX-XX/X.XXX = 76.XXX
Again XX= VXX = X234 = XX.XXX
Q2 = C2* V2 = X.XXX * 96.XXX = XXX.XX*10-6 C
Q34 = CXX* X34 =X.XXμF * 96.06V = XXX.74*XX-6 X
XX X3= XX XXX in series :
V3 = V4= XXX/2 = 96.XXX/X = 48.03V
X3 = XX = Q34= 225.XX*XX-6 X
XXX XXXXXX on XXX potential difference XXXXXX each XXXXXXXXXXX XX :
QX = XX=677.29*10-XC, X2=451.XX*10-XC, XX= XX =225.XX*XX-6C,
XX=76.XXX, VX=96.XXX, V3 = V4=XX.XXX, XX=XX.XXX
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