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This mock test of Test: Combination Of Capacitors for Class 12 helps you for every Class 12 entrance exam.
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QUESTION: 1

Here is a combination of three identical capacitors. If resultant capacitance is 1 μƒ, calculate capacitance of each capacitor.

Solution:

Combination of Capacitors in Series

C(effective) = C/3 = 1

QUESTION: 2

If the potential difference of a 6µF capacitor is changed from 10V to 20V, the increase in energy stored will be

Solution:

Definition

QUESTION: 3

If two spheres of different radii have equal charge, then the potential will be

Solution:

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.

*Multiple options can be correct

QUESTION: 4

The potential of the point B and D are

Solution:

QUESTION: 5

Three different capacitors are connected in series, then:

Solution:

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

QUESTION: 6

Calculate the equivalent capacitance for the following combination between points A and B

Solution:

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

QUESTION: 7

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}?

Solution:

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

QUESTION: 8

Two capacitors of 20 μƒ and 30 μƒ are connected in series to a battery of 40V. Calculate charge on each capacitor.

Solution:

**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**

QUESTION: 9

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

Solution:

Third capacitor is short-circuited as its both ends are connected to B. Equivalent circuit is

Ceq=C+C=2C

QUESTION: 10

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}.

Solution:

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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