In the network shown above, determine the potential difference across the 5µF capacitor
3V
6V
12V
18V
Explanation
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c=15×5/15+5
c=3.75×10^-6
c=q/v
q=24×3.75
q=90×10^-6
v=q/v
v=90×10^-6/5×10^-6
v=18v
5µC and 15µC are series, C = (5×15)/(5+15) = 75/20 = 3.75
Charge following through the series capacitors is
Q = CV = 3.75 ×24 = 90µC
The same charge (90µC)) flow through 5µC
potential difference across 5µC, V= Q/C = 90µC/5µC = 18V.
In series, the same charge (Q) flows, so voltage divides inversely to capacitance:
V5uf = 15/(5 + 15} ×24 = 18v
V15uf = {5}/{20}× 24 = 6V
bcs they are inversely proportional
Q=CV
V= 24
C=1/5 + 1/15(FIND LCM AND SOLVE)
= 3+1/15
1/C =4/15
C. = 15/14 =3.75
Q= 3.75x24
=90
FOR THE 5uf
Q= CV
90=5V
V=90/5
=18(ANS)
c=15×5/15+5
c=3.75×10^-6
c=q/v
q=24×3.75
q=90×10^-6
v=q/v
v=90×10^-6/5×10^-6
v=18v
i understand this one more
C1=5uF. C2=16uF(series)V=24
Potential difference=V×c2/c1+c2
P.d=24×(15/5+15)
P.d=24×(15/20)
P.d=24×3/4
P.d=72/4
P.d=18v
