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Multiple choice physics capacitance capacitors in parallel combination of capacitors capacitors in parallel and series

A capacitors of $2 \mu F$ is required is an electric circuit across a potential difference of 1.0kv. A large number of $1 \mu F$ capacitors are available which can with stand a potential difference of not more than 300V The minimum number of capacitors required to achieve this is:

  1. 24

  2. 32

  3. 2

  4. 16
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

To withstand 1000V using 300V capacitors, we need at least 4 in series (4 * 300 = 1200V). To get 2uF total from 1uF capacitors, we need 2 parallel branches of 4 capacitors each, totaling 8 capacitors. However, the calculation for 1000V/300V requires 4 in series, and to get 2uF from 1uF (where each series branch is 0.25uF), we need 8 branches. 8 * 4 = 32.

Multiple choice physics capacitance capacitors in parallel combination of capacitors capacitors in parallel and series

Two capacitors were charged to potentials 80 and 30 V. Then they connected in parallel. The potential difference across both condensers is 60 V. The ratio of the capacitances of the capacitors is

  1. 3 : 2

  2. 1 : 3

  3. 1 : 4

  4. 2 : 3
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using charge conservation, Q1 + Q2 = (C1 + C2) * V_common. C1(80) + C2(30) = (C1 + C2) * 60. 80C1 + 30C2 = 60C1 + 60C2. 20C1 = 30C2. C1/C2 = 30/20 = 3/2.

Multiple choice physics capacitance capacitors in parallel combination of capacitors capacitors in parallel and series

The plates of a parallel plate capacitor are $4$cm apart, the first plate is at $300$V and the second plate at $-100$V. The voltage at $3$cm from the second plate is?

  1. $200$V
  2. $400$V
  3. $250$V
  4. $500$V
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The potential varies linearly between plates. Plate 1 is at 300V (x=0), Plate 2 is at -100V (x=4cm). The potential at distance x from Plate 1 is V(x) = 300 - (300 - (-100))/4 * x = 300 - 100x. At 3cm from Plate 2 (which is 1cm from Plate 1), V = 300 - 100(1) = 200V.

Multiple choice physics capacitance capacitors in parallel combination of capacitors capacitors in parallel and series

Find the total capacitance for three capacitors of $10$f,$15$f and $35$f in parallel with each other?

  1. $20f$
  2. $50f$
  3. $60f$
  4. $10f$
  5. $5f$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given :    $C _1 = 10$ f                  $C _2 = 15$  f                    $C _3 = 35$  f

Equivalent capacitance for parallel combination          $C _{eq} = C _1 + C _2 + C _3$
$\therefore$   $C _{eq} = 10 + 15 + 35  = 60$  $f$

Multiple choice physics capacitance capacitors in parallel combination of capacitors capacitors in parallel and series

Two capacitors of capacity $C _1$ and $C _2$ are connected in parallel, then the equivalent capacity is:

  1. $C _1+C _2$
  2. $C _1C _2/(C _1+C _2)$
  3. $C _1/C _2$
  4. $C _2/C _1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$C _1, C _2$ are connected in parallel then equivalent capacitance is calculated as
$V=V _1=V _2$.....(1)
$q=q _1+q _2$
$\therefore CV=C _1V _1+C _2V _2$
From (1) $C=C _1+C _2$

Multiple choice physics capacitance capacitors in parallel combination of capacitors capacitors in parallel and series

Capacity of a parallel plate capacitor is $2\mu F$. The two plates of the capacitor are given $400\mu C$ and $-200\mu C$charges respectively. The potential difference between the plates is 

  1. $100\ V$
  2. $200\ V$
  3. $300\ V$
  4. $150\ V$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The potential difference between plates is V = (Q1 - Q2) / (2C) is incorrect. The potential difference is determined by the charge on the inner surfaces. For a capacitor with charges Q1 and Q2, the charge on the inner faces is (Q1 - Q2)/2. Here, (400 - (-200))/2 = 300uC. V = Q/C = 300uC / 2uF = 150V.

Multiple choice physics capacitance capacitors in parallel combination of capacitors capacitors in parallel and series

A parallel plate capacitor consist of two circular plates each of radius 2 cm, separated by a distance of 0.1 mm. If voltage across the plates is varying at the rate of $5 \times {10^{13}}V{s^{ - 1}}$ , then the value of displacement current is:

  1. $5.50A$
  2. $ 5.56 \times 10^2 A $
  3. $ 5.56 \times 10^3 A $
  4. $ 2.28 \times 10^4 A $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Displacement current Id = e0 * d(Phi_E)/dt = e0 * A * dE/dt = e0 * A * (1/d) * dV/dt = (e0 * A / d) * dV/dt = C * dV/dt. C = e0 * pi * r^2 / d. C = (8.85e-12 * 3.14 * 0.02^2) / 0.0001 = 1.11e-10 F. Id = 1.11e-10 * 5e13 = 5.55 A.

Multiple choice physics capacitance capacitors in parallel combination of capacitors capacitors in parallel and series

When $n$ identical capacitors are connected in series their effective capacity is $C _s$ and when they are connected in parallel their effective capacity is $C _p$. The relation between $C _p$ and $C _s$ is:

  1. $C _p = n \,C _s$
  2. $C _p = \dfrac{C _s}{n}$
  3. $C _p = n^2 \,C _s$
  4. $C _p = \dfrac{C _s}{n^2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For n identical capacitors of capacity C, Cs = C/n and Cp = nC. Therefore, Cp = n * (n * Cs) = n^2 * Cs.