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Here is a combination of three identical capacitors. If resultant capacitance is 1 μƒ, calculate capacitance of each capacitor.
Combination of Capacitors in Series
C(effective) = C/3 = 1
If the potential difference of a 6µF capacitor is changed from 10V to 20V, the increase in energy stored will be
Definition
If two spheres of different radii have equal charge, then the potential will be
When equal charges are given to two spheres of different radii, the potential will be more or the smaller sphere as per the equation, Potential = Charge / Radius.
Since potential is inversely proportional to radius, the smaller radius will have higher potential and vice versa.
The capacitor preferred for highfrequency circuit is
Mica capacitor. Mica capacitors have low resistive and inductive components associated with it. Hence, they have high Q factor and because of high Q factor their characteristics are mostly frequency independent, which allows this capacitor to work at high frequency.
Three different capacitors are connected in series, then:
C = Q/V
series connection splits the battery potential, hence....
V = V_{1} + V_{2}
V_{1} = Q/C_{1} & V_{2} = Q/C_{2}
Therefore, both have equal charge
Calculate the equivalent capacitance for the following combination between points A and B
4μf and 6μf are in series (right side of AB) so 6x(4/6)+4=24/10μf
And now 5 μf and the resulting of above two are in parallel (as on different sides of AB) so 5+(24/10) =50+(24/10)=74/10=37/5μf
There are three capacitors with equal capacitance. In series combination, they have a net capacitance of C_{1 }and in parallel combination, a net capacitance of C_{2}.What will be the value of C_{1} C_{2}?
For series combination,
1/C1=(1/C)+(1/C)+(1/C) [ Since all capacitors have equal capacitances]=3/C Or
C1=C/3−−−(i)
For parallel combination,
C2=C+C+C=3C−−−(ii)
Now,C1/C2=(C/3)/3C=(C/3)×(1/3C)=1/9
Two capacitors of 20 μƒ and 30 μƒ are connected in series to a battery of 40V. Calculate charge on each capacitor.
C1= 20×10µf
and C2= 30×10µf
in series Ceq = C1C2/(C1+C2)
Ceq = 20×10^(6)×30×10^(6)/20×10^(6)+30^×10(6)
Ceq= 12×10^(6)f
As we know that Q = CV
Putting the values of C and V= 40V, we get
Q = (12 * 10^6) * 40
= 480µC
Two capacitors of equal capacity are first connected in parallel and then in series. The ratio of the total capacities in the two cases will be
Third capacitor is shortcircuited as its both ends are connected to B. Equivalent circuit is
Ceq=C+C=2C
When two capacitors C_{1} and C_{2} are connected in series and parallel, their equivalent capacitances comes out to be 3μƒ and 16μƒ respectively. Calculate values of C_{1} and C_{2}.
Let C_{p} be the equivalent capacitance of parallel
C_{p}=C_{1}+C_{2}
16= C_{1}+C_{2}……..(1)
Let Cs be the equivalent of capacitance in series
1/C_{s}=(1/C_{1}) + (1/C_{2})
Or, C_{s}=C_{1}C_{2}/C_{1}+C_{2}
Or,3=C_{1}C_{2}/16
Or,C_{2}C_{1}=48
C_{2}=48/C_{1}……..(ii)
16=C1+48/C1
(C_{1}^{2}16C_{1}+48)=0
(C_{1}4)(C_{1}12)=0
C_{1}=4 ; C_{1}=12
If,
C_{1}=4
C_{2}=12
If,
C_{2}=4
C_{1}=12
So, the answer is, Either 12μf or4μf
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