The following question appeared in Karnataka Common Entrance Test (CET) 2010 question paper. Even though at the first reading it may appear somewhat difficult for you on seeing the circuit, a careful look at the circuit will assure you that it’s simple.
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiirFVD5P-XDIzQWAX4opvS6nPkFZ4qmnGQBicQSU1DRGMa-J_lN2kpOQqE0Fdceq7XqF_0W59pXByg4SoCGUbR0do7YKpcokWoNY6QPpdNO_In6s8rjL998oE2hdo4bLlBlGCT/s320/Capacitor-pp9-7-11.jpg)
All capacitors used in the diagram are identical and each is of capacitance C. Then the effective capacitance between the points A and B is
(a) 1.5 C
(b) 6 C
(c) C
(d) 3 C
The first three capacitors are in parallel, giving an effective capacitance 3 C. The last three capacitors also are in parallel, giving an effective capacitance 3 C.. These two parallel combinations are connected in series. Therefore, the effective capacitance between the points A and B is 3 C/2 = 1.5 C.
The following question appeared in the Karnataka CET 2008 question paper:
How many 6 μF, 200 V condensers are needed to make a condenser of 18 μF, 600 V?
(1) 9
(2) 18
(3) 3
(4) 27
Since the voltage rating of each 6 μF capacitor is 200 V, you should connect 3 capacitors in series to obtain the required voltage rating of 600 V. But then the effective capacitance of the series combination of these three capacitors will be 2 μF (6/3 = 2). To obtain the required capacitance of 18 μF, you need to connect nine such series combinations in parallel. So the total number of capacitors required is 9×3 = 27.
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgO1VKRwOMslrYGUSWQSGNplAa2w0aLy7Jh5smSaJALVjwu6RSZP6iXp48PYr3WtxeKxXKTc78UbZ-2x1MDOT1TH_2sNm95XMfWPBs7SC2bBQZSEDuaBR4grO4zuu0tqE3RJzO1/s320/Capacitor1-pp9-7-11.jpg)
The following question also appeared in the Karnataka CET 2008 question paper:
The total energy stored in the condenser system shown in the figure will be
(1) 2 μJ
(2) 4 μJ
(3) 8 μJ
(4) 16 μJ
The series combination of 6 μF and 3 μF gives an effective capacitance of (6×3)/(6+3) = 2 μF. Since this is in parallel with a 2 μF capacitor, the effective capacitance C across the 2 volt battery is 4 μF. The energy stored in the system of capacitors is ½ CV2 = (½)×(4×10–6) ×22 = 8×10–6 J = 8 μJ.
hv film capacitor manufacturer Thanks for a very interesting blog. What else may I get that kind of info written in such a perfect approach? I’ve a undertaking that I am simply now operating on, and I have been at the look out for such info.
ReplyDelete