Tag: capacitors in parallel and series

Quizzes (0) Questions (70)

Questions Related to capacitors in parallel and series

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

$4\ \mu F$ and $6\ \mu F$ capacitors are joined in series and $500\ v$ are applied between the outer plates of the system. What is the charge on each plate ?

  1. $1\cdot 2\times 10^{3}\ C$

  2. $6\cdot 0\times 10^{3}\ C$

  3. $5\cdot 0\times 10^{-3}\ C$

  4. $2\cdot 0\times 10^{-3}\ C$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Since the capacities are connected in series

$\dfrac{1}{C _{eq}}=\dfrac{1}{C _1}+\dfrac{1}{C _2}$
$\dfrac{1}{C _{eq}}=\dfrac{1}{4\mu}+\dfrac{1}{6\mu}$
$C _{eq}=\dfrac{12\mu}{5}$
$C=\dfrac{Q}{V}$
$\dfrac{12\mu}{5}=\dfrac{Q}{500}$
$Q=1.2\times 10^{3}$

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

A parallel plate capacitor has two square plates with equal and opposite charges. The surface charge densities on the plate are $+\sigma$ and $-\sigma$ respectively. In the region between the plates the magnitude of electric field is:

  1. $\dfrac { \sigma }{ 2{ \varepsilon } _{ 0 } } $

  2. $\dfrac { \sigma }{ { \varepsilon } _{ 0 } } $

  3. 0

  4. none of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The magnitude of the electric field due to a charged plate is given as:

$\vec E=\dfrac{\sigma}{2\varepsilon _0}$

The charge density on the upper plate is positive whereas on the lower plate it is negative.

Therefore, the electric field in the region between the two plates is the difference of the two fields.
It is given as:
$E _{net}=\dfrac{\sigma}{2\varepsilon _0}-\dfrac{-\sigma}{2\varepsilon _0}$

$E _{net}=\dfrac{\sigma}{\varepsilon _0}$

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.