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

\${ C }_{ P }=2+4=6\mu F\$\$\cfrac { 1 }{ C } =\cfrac { 1 }{ 6 } +\cfrac { 1 }{ 6 } =\cfrac { 2 }{ 6 } =\cfrac { 1 }{ 3 } \quad or\quad C=3\mu F\$Total charge\$Q=CV=3\times 12=36\mu C\quad \quad \$voltage across \$6\mu F\$ capacitor \$\quad =\cfrac { 36\mu C }{ 6\mu F } =6V\$\$\therefore\$ Voltage across each of \$2\mu F\$ and \$4\mu F\$ capacitors \$=12V-6V\$\$V=6V\$

Two capacitors when connected in series have a capacitance of \$3\mu F\$, and when connected in parallel have a capacitance of \$16\mu F\$. Their individual capacities are :

Let their individual capacities are \$C_1\$ and \$C_2\$.Here, \$C_S=\dfrac{C_1C_2}{C_1+C_2}=3 ...(1)\$ and \$C_P=C_1+C_2=16 ...(2)\$From (1) and (2), \$C_1C_2=3(16)=48 ...(3)\$From (2), (3), \$C_1+48/C_1=16 \Rightarrow C_1^2-16C_1+48=0\$or \$C_1^2-12C_1-4C_2+48=0\$or \$C_1(C_1-12)-4(C_1-12)=0\$or \$(C_1-12)(C_1-4)=0\$\$C_1=12, 4 \mu F\$From (3), for \$C_1=12, C_2=\dfrac{48}{12}=4 \mu F\$ and for \$C_1=4, C_2=\dfrac{48}{4}=12 \mu F\$

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>>Class 12
>>Physics
>>Electrostatic Potential and Capacitance
>>Combination of Capacitors
>> $Parallel Combination of Capacitors$

Parallel Combination of Capacitors

Q: In a parallel plate capacitor with air between the plates, each plate has an area of \$6\times 10^{-3}m^2\$ and the distance between the plates is 3mm. Calculate the capacitance of the capacitor. If this capacitor is connected to a 100 V supply, what is the charge on each plate of the capacitor?

Area of each plate \$A= 6\times 10^{-3} m^2\$Distance between the plates \$d = 3mm = 0.003 m\$Capacitance of capacitor \$C = \dfrac{A\epsilon_o}{d}\$\$\therefore\$ \$C = \dfrac{6\times 10^{-3} \times 8.85\times 10^{-12}}{0.003} = 17.7 pF\$Voltage across the capacitor \$V =100\$ voltsThus charge on each plate \$Q = CV\$\$\therefore\$ \$Q = 17.7\times 10^{-12} \times 100\$\$\implies\$ \$Q = 1.77 \times 10^{-9}\$ C

Q: Three capacitors of capacitances 2 pF, 3 pF and 4 pF are connected in parallel.(a) What is the total capacitance of the combination?(b) Determine the charge on each capacitor if the combination is connected to a 100 V supply.

(a) For parallel combination, \$C_{eq}=C_{1}+C_{2}+C_{3}=(2+3+4)pF =9pF \$(b) \$q=V\times C\$for parallel array, V remains the same across each capacitor.for \$C=2\ pF\$\$q_1=VC=100\times 2\times 10^{-12}=2\times 10^{-10}C\$for \$C=3\ pF\$\$q_2=VC=100\times 3\times 10^{-12}=3\times 10^{-10}C\$for \$C=4\ pF\$\$q_1=VC=100\times 4\times 10^{-12}=4\times 10^{-10}C\$

Q: A parallel plate air capacitor has a capacitance \$C\$. When it is half filled with a dielectric of dielectric constant \$5\$, the percentage increase in the capacitance will be

Initial capacitance\$=\dfrac{{\epsilon}_{\circ}A}{d}\$When it is half filled by a dielectric of dielectric constant \$k\$ then \${C}_{1}=\dfrac{k{\epsilon}_{\circ}A}{\frac{d}{2}}=2k\dfrac{{\epsilon}_{\circ}A}{d}\$and \${C}_{2}=\dfrac{{\epsilon}_{\circ}A}{\frac{d}{2}}=\dfrac{2{\epsilon}_{\circ}A}{d}\$\$\dfrac{1}{C}=\dfrac{1}{{C}_{1}}+\dfrac{1}{{C}_{2}}\$\$=\dfrac{d}{2{\epsilon}_{\circ}A}\left(\dfrac{1}{k}+1\right)\$\$=\dfrac{d}{2{\epsilon}_{\circ}A}\left(\dfrac{1}{5}+1\right)\$\$=\dfrac{d}{2{\epsilon}_{\circ}A}\left(\dfrac{6}{5}\right)\$\$=\dfrac{3d}{5{\epsilon}_{\circ}A}\$\$\therefore C=\dfrac{5{\epsilon}_{\circ}A}{3d}\$Hence, \$\%\$ increase in capacitance is\$=\left(\dfrac{\dfrac{5{\epsilon}_{\circ}A}{d}-\dfrac{{\epsilon}_{\circ}A}{d}}{\dfrac{{\epsilon}_{\circ}A}{d}}\right)\times 100=\dfrac{2}{3}\times 100=66.6\%\$

Q: Three capacitors each of capacitance \$9\ pF\$ are connected in series as shown in figure.(a) What is the total capacitance of the combination?(b) What is the potential difference across each capacitor, if the combination is connected to a \$120\ volt\$ supply?

\$(A)\$ Effective capacitance in series combination is given by:\$\cfrac{1}{C_s}=\cfrac{1}{C_1}+\cfrac{1}{C_2}+\cfrac{1}{C_3}\$\$\cfrac{1}{C_s}=\cfrac{1}{9}+\cfrac{1}{9}+\cfrac{1}{9}\$\$\Rightarrow C_s= 03 pF\$so, total capacitance of the combination is \$3 pF\$\$(B)\$ \$Q=C_sV\$\$Q=3\times 120=360pC\$\$V_1=\cfrac{Q}{C_1}\$ (as in series combination, charges is constant)\$\Rightarrow\cfrac{360}{9}=40V\$\$V_2=\cfrac{Q}{C_2}=\cfrac{360}{9}=40 V\$\$V_3=\cfrac{Q}{C_3}=\cfrac{360}{9}=40V\$

Q: Derive an expression for effective capacitance of three capacitors connected in parallel.

In the given figure, capacitors \${{C}_{1}},{{C}_{2}}\,and\,\,{{C}_{3}}\$are connected in parallel. The same potential is applied across all the three capacitors. Let \${{Q}_{1}},{{Q}_{2}}\,and\,{{Q}_{3}}\$are charges on the three plates of capacitors such that \$ {{Q}_{1}}={{C}_{1}}V\, \$ \$ {{Q}_{2}}={{C}_{2}}V \$ \$ {{Q}_{3}}={{C}_{3}}V \$ The total charge stored is \$Q={{Q}_{1}}+{{Q}_{2}}+{{Q}_{3}}\$ and equivalent capacitance is \${{C}_{eq}}=\dfrac{Q}{V}\$ So, \$ {{C}_{eq}}=\dfrac{{{Q}_{1}}+{{Q}_{2}}+{{Q}_{3}}}{V} \$ \$ =\dfrac{{{Q}_{1}}}{V}+\dfrac{{{Q}_{2}}}{V}+\dfrac{{{Q}_{3}}}{V} \$ \$ {{C}_{eq}}={{C}_{1}}+{{C}_{2}}+{{C}_{3}} \$

Q: Find equivalent capacitance between points A and B :

Hint:If Two Capacitors \$\bf{C_1}\$ and \$\bf{C_2}\$ are connected in series then, their equivalent capacitance is given by\$\bf{C_{eq.} = \dfrac{C_1 .C_2}{C_1+C_2}}\$If Two Capacitors \$C_1\$ and \$C_2\$ are connected in parallel then, their equivalent capacitance is given by\$\bf{C_{eq.} = C_1 + C_2}\$\$\textbf{Correct option: Option (B).}\$ \$\textbf{Explanation:}\$ \$\textbf{Step 1: Drawing the equivalent circuit and finding capacitance in the upper arm.}\$ The figure shows the equivalent circuit for the given circuit. The capacitors in the upper arm are in series so the total capacitance in the upper arm will be given by, \${{C}_{u}}={{\left( \dfrac{1}{C}+\dfrac{1}{C}+\dfrac{1}{C} \right)}^{-1}}={{\left( \dfrac{3}{C} \right)}^{-1}}=\dfrac{C}{3}\$ \$\textbf{Step 2: Finding the total capacitance.}\$ Now the equivalent capacitor in the upper arm and the bottom capacitor are in parallel so the equivalent capacitance of the circuit will be, \${{C}_{eq}}=\dfrac{C}{3}+C=\dfrac{4C}{3}\$ \$\textbf{Thus, the equivalent capacitance of the circuit is \dfrac{4C}{3} . Hence, option (B) is correct.}\$

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