Electrostatic Potential and Capacitance Test

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Electrostatic Potential and Capacitance Test - 30

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Question 1
1 / -0

The potential difference across $$3 \mu F$$ condenser is :

A
$$40 Volt$$

B
$$60 Volt$$

C
$$80 Volt$$

D
$$120 Volt$$

Solution

As both the capacitor are in series the charge flowing through it is same,
$$Q_1=Q_2$$
Now if we consider potential across $$3\mu F$$ capacitor as V then potential across $$6\mu F$$ capacitor is given as $$120-V$$ ,
$$(120-V)6\mu F=V(3\mu F)$$
$$720 \mu F =V(9\mu C) $$
$$V = 80 V$$
Question 2
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Two condensers of capacitance $$4\mu F$$ and $$5\mu F$$ are joined in series. If the potential difference across $$5\mu F$$ is $$10V$$, then the potential difference across $$4\mu F$$ condenser is :

A
$$22.5V$$

B
$$10V$$

C
$$12.5V$$

D
$$25V$$

Solution

When two capacitors are in series $$Q$$ present on then is same
$$\therefore\ Q\ on \ 5\mu F$$ is $$Q=5\times10^{-6}\times 10$$
$$Q=5\times 10^{-5}$$
$$\therefore Q$$ on $$4\mu F$$ is $$5\times 10^{-5}$$
$$\therefore V=\dfrac{Q}{C}$$
$$V=\dfrac{5\times 10^{-5}}{4\times 10^{-6}}$$
$$V=12.5\ V$$
Question 3
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Charge Q taken from the battery of 12V in the circuit is :

A
$$72 \mu C$$

B
$$36 \mu C$$

C
$$156 \mu C$$

D
$$20 \mu C$$

Solution

The equivalent capacitance of the circuit is ,
$$C=4+\dfrac{3\times 6}{3+6}=4+2=6 \mu F$$
Thus, Charge taken from the battery is,
$$Q=CV=6\times 12=72 \mu C$$
Question 4
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The equivalent capacitance of the network given below is$$1 \mu F$$. The value of C is:

A
$$3 \mu F$$

B
$$1.5 \mu F$$

C
$$2.5 \mu F$$

D
$$1 \mu F$$

Solution
Question 5
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Three capacitors $$2\mu F, 3\mu F$$ and $$5\mu F$$ are connected in parallel. The capacitance of the combination:

A
$$\dfrac{30}{31}\mu F$$

B
$$\dfrac{31}{30}\mu F$$

C
$$ 10 \mu F$$

D
$$2.5 \mu F$$

Solution

When connected in parallel ,
$$C_{eff}=C_{1}+C_{2}+C_{3}$$
$$C_{eff}=2+3+5$$
$$C_{eff}=10\mu F$$
Question 6
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The ratio of the resultant capacities when three capacitors of $$2 \mu F , 4 \mu F$$ and $$6 \mu F$$ are connected first in series and then in parallel is

A
$$1 : 11$$

B
$$11 : 1$$

C
$$12 : 1$$

D
$$1 : 12$$

Solution

When capacitor in series
$$\dfrac{1}{C_{eff_s}}=\dfrac{1}{C_{1}}+\dfrac{1}{C_{2}}+\dfrac{1}{C_{3}}$$

$$\Rightarrow \dfrac{1}{C_{eff}}=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}$$

$$\Rightarrow \dfrac{1}{C_{eff}}=\dfrac{6+3+2}{12}$$

$$\Rightarrow{C_{eff}}=\dfrac{12}{11}$$

$$C_{{eff}_{p}}=C_{1}+C_{2}+C_{3}$$

$$=2+4+6$$

$$=12$$

$$\therefore \dfrac{C_{S}}{C_{P}}=\dfrac{1}{11}$$
Question 7
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The maximum and minimum resultant capacity that can be obtained with $$2\mu F, 3\mu F$$ and $$6\mu F$$ are respectively:

A
$$11\mu F, 1\mu F$$

B
$$11\mu F, 6\mu F$$

C
$$11\mu F, 2\mu F$$

D
$$11\mu F, 4\mu F$$

Solution

Maximum capacitors is obtained when they are connected in parallel
$$C_{eff}=C_{1}+C_{2}+C_{3}$$
$$C_{max}=2+3+6$$
$$C_{max}=11\mu F$$
Minimum capacitance occurs when capacitors are connected in series. So:
$$\dfrac{1}{C_{min}}=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{6}$$
or, $$C_{min}=1\mu F$$
Question 8
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Three capacitors $$2\mu F, 3\mu F$$ and $$6\mu F$$ are connected in series. The effective capacitance of the combination is:

A
$$11\ \mu F$$

B
$$1\ \mu F$$

C
$$1.2\ \mu F$$

D
$$2\ \mu F$$

Solution

When capacitors are in series
$$\dfrac{1}{C_{eff}}=\dfrac{1}{C_{1}}+\dfrac{1}{C_{2}}+\dfrac{1}{C_{3}}$$$$=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{6}$$

$$\Rightarrow C_{eff}=1\ \mu F$$
Question 9
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A parallel plate condenser has initially air medium between the plates. If a slab of dielectric constant 5 having thickness half the difference of separation between the plates is introduced, the percentage increase in its capacity is :

A
33.3%

B
66.7%

C
50%

D
75%

Solution

$$t=\dfrac{d}{2}, K= 25$$
$$C=\dfrac { { \varepsilon }_{ o }A }{ \left[ \left( d-t \right) +\dfrac { t }{ k } \right] } $$
$$C=\dfrac { { \varepsilon }_{ o }A }{ d } $$
Initially capacitance is C
$${ C }^{ 1 }=\dfrac { { \varepsilon }_{ o }A }{ \left[ d-\dfrac { d }{ 2 } +\dfrac { d }{ 2\left( 5 \right) } \right] } $$
$${ C }^{ 1 }=\dfrac { { \varepsilon }_{ o }A }{ \dfrac { d }{ 2 } \left[ 1+\dfrac { 1 }{ 5 } \right] } $$
$${ C }^{ 1 }=\dfrac { { \varepsilon }_{ o }A }{ d\left[ 1+\dfrac { 1 }{ 5 } \right] } $$
$${ C }^{ 1 }=\dfrac { { \varepsilon }_{ o }A }{ d\left[ 1+\dfrac { 1 }{ 5 } \right] } $$
Therefore % increase =$$\left[ \dfrac { 10 }{ 6 } -1 \right] \times 100=66.7\%$$
Question 10
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Two identical metal plates separated by a distance $$d$$ forms parallel plate capacitor of capacity $$C$$. A metal sheet of thickness $$d/2$$ and same dimensions is inserted between the plates, so that the air gap is separated into two equal parts. The new capacity of the system will be:

A
$$\displaystyle \dfrac{\varepsilon _{0}A}{2d}$$

B
$$\displaystyle \frac{\varepsilon _{0}A}{4d}$$

C
$$\displaystyle \frac{2\epsilon_{0}A}{d}$$

D
$$\displaystyle \frac{\epsilon_{0}A}{d}$$

Solution

The final capacitance in series combination of
$$\dfrac{ \varepsilon _0 \ A}{\dfrac{d}{4}}$$ and $$\dfrac{ \varepsilon _0\ A}{\dfrac{d}{4}}$$ is
$$\dfrac{1}{C} = \dfrac{2d}{4\epsilon_0 A } $$
Hence , $$ C = \dfrac{2\epsilon_0 A }{d}$$