Electrostatic Potential and Capacitance Test

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Electrostatic Potential and Capacitance Test - 7

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Weekly Quiz Competition

Question 1
4 / -1

Three capacitors each of capacity 4 μF are to be connected in such a way that the effective capacitance is 6 μF. This can be done by

A
connecting all of them in series

B
connecting them in parallel

C
connecting two in series and one in parallel

D
connecting two in parallel and one in series

Solution

To get equivalent capacitance 6 μF. Out of the 4 μF capacitance, two are connected in series and third one is connected in parallel.
Question 2
4 / -1

The equivalent capacitance for the network shown in the figure is

A

B

C

D

Solution

Capacitance of C₁= C₄ = 100pF
Capacitance of C₂ = C₃ = 400pF
Supply voltage , V = 400V
Capacitors C₂ and C₃ are connected is series,
Equivalent capacitance
Capacitors C₁ and C are in parallel
Their equivalent capacitance.
C = C + C₁ = 200 + 100 = 300pF
Capacitors C'' and C₄ are connected in series Equivalent capacitance
Question 3
4 / -1

A network of four 20μF capacitors is connected to a 600V supply as shown in the figure.

The equivalent capacitance of the network is

A
30.26 μF

B
20 μF

C
26.67 μF

D
10 μF

Solution

Form the figure C₁, C₂, C₃ are connected in series.

Now C_s is in parallel with C₄
∴ equivalent capacitance C_eq = C_s + C₄
Question 4
4 / -1

The charge on capacitors C₁ and C₄ are

A
4 × 10⁻³C, 12 × 10⁻³C

B
6 × 10⁻³C, 12 × 10⁻³C

C
2 × 10⁻³C, 4 × 10⁻³C

D
3 × 10⁻³C, 2 × 10⁻³C

Solution

Charge on each capacitor C₁, C₂, C₃ is same = q
Let charge on C₄ be q.

Also qC₄ = 600
or q = C₄ × 600μC = 20 × 600μC = 12 × 10⁻³C
Question 5
4 / -1

The charge on 3μF capacitor shown in the figure is

A
2 μC

B
10 μC

C
6 μC

D
8 μC

Solution

Since all the capacitors are connected in series,
∴ equivalent capacitance is

Charge on each capacitor is same (∵ V = 10V)
q = C_eqV = 1 × 10 = 10μC
Question 6
4 / -1

Minimum number of capacitors each of 8μF and 250V used to make a composite capacitor of 16μF and 1000V are

A
8

B
32

C
16

D
24

Solution

Minimum number of capacitors in each row = 1000/250 = 4
Therefore, 4 capacitors connected in series.
If there are m such rows, that total capacity
= m × 2 = 16 ∴ m = 16/28
∴ minimum number of capacitors = 4 × 8 = 32
Question 7
4 / -1

A capacitor of capacitance C₁ is charged to a potential V and then connected in parallel to an uncharged capacitor of capacitance C₂. The final potential difference across each capacitor will be

A

B

C

D

Solution

When a charged capacitor of capacitance C₁ is connected in parallel to an uncharged capacitor of capacitance C₂, the charge is shared between them till both attain the common potential which is given by, common Potential = q₁ + q₂/C₁+ C₂ = q₁ + q₂/C₁ + C₂ = C₁V/C₁ + C₂
Question 8
4 / -1

Two capacitors of 2 μF and 4 μF are connected in parallel. A third capacitor of 6 μF is connected in series. The combination is connected across a 12 V battery. The voltage across 2 μF capacitor is

A
2 V

B
8 V

C
6 V

D
1 V

Solution

C_p = 2 + 4 = 6μF

Total Q = CV = 3 × 12 = 36μF
Voltage across 6μF capacitor = 36μC/6μF = 6V
∴ Voltage across each of 2μF and 4μF capacitors = 12V - 6V
V = 6V
Question 9
4 / -1

Two identical capacitors are joined in parallel, charged to a potential V, separated and then connected in series, the positive plate of one is connected to the negative of the other. Which of the following is true?

A
The charges on the free plated connected together are destroyed.

B
The energy stored in the system increases.

C
The potential difference between the free plates is 2 V.

D
The potential difference remains constant.

Solution

When the two capacitors charged to same potential are connected in series, then total potential difference V′ = V₁ + V₂ = V + V = 2V
Question 10
4 / -1

In the given circuit, initially K₁ is closed and K₂ is open. Then K₁ is opened and K₂ is closed. If q₁′ and q₂′ are charges on C₁ and C₂ and V₁ and V₂ are the voltage respectively, then

A
charge on C₁ gets redistributed such that V₁ = V₂

B
charge on C₁ gets redistributed such that q₁′ = q₂′

C
charge on C₁ gets redistributed such that C₁V₁ = C₂V₂ = C₁V

D
charge on C₁ gets redistributed such that q₁′ + q₂′ = 2q

Solution

From the figure, when K₁ is closed and K₂ is open, C₁ is charged to potential V acquiring a total charge q = C₁V When K₁ is open and K₂ is closed, battery is cut off C₁ and C₂ are in parallel. The charge on C₁ is shared between C₁ and C₂ such that V₁ = V₂
As there is no loss of charge q₁′ + q₂′ = q