166028
Two capacitors of capacitance $C$ are connected in series If one of them is filled with dielectric substance $k$, what is the effective capacitance ?
1 $\frac{\mathrm{kC}}{(1+\mathrm{k})}$
2 $\mathrm{C}(\mathrm{k}+1)$
3 $\frac{2 \mathrm{kC}}{1+\mathrm{k}}$
4 none of these
Explanation:
: When plates of Capacitor are separated by a dielectric Medium of dielectric Constant k, its capacity $\mathrm{C}_{\mathrm{m}}=\frac{\mathrm{k} \varepsilon_{0} \mathrm{~A}}{\mathrm{~d}}=\mathrm{kC}_{0}=\mathrm{kC} \quad\left[\right.$ Where, $\left.\mathrm{C}_{0}=\mathrm{C}\right]$ Now, two capacitor $\mathrm{kC}$ and $\mathrm{C}$ are in Series, that effect Capacitance, $\frac{1}{\mathrm{C}}=\frac{1}{\mathrm{kC}}+\frac{1}{\mathrm{C}}$ $\Rightarrow \frac{1}{\mathrm{C}^{\prime}}=\frac{1+\mathrm{k}}{\mathrm{kC}}$ $\Rightarrow \mathrm{C}^{\prime}=\frac{\mathrm{kC}}{\mathrm{k}+1}$
[DCE-2007]
Capacitance
166030
In a parallel plate air capacitor of capacitance $4 \mathrm{~F}$ if the lower half of air space is filled with a material of dielectric constant 3 , its capacitance changes to
1 $\frac{4}{3} \mathrm{~F}$
2 $\frac{8}{3} \mathrm{~F}$
3 $8 \mathrm{~F}$
4 $12 \mathrm{~F}$
Explanation:
: Net capacitance, $\mathrm{C}=\mathrm{C}_{1}+\mathrm{C}_{2} \quad \ldots$ (i) Given, air capacitance $\mathrm{C}=\frac{\varepsilon_{0} \mathrm{~A}}{\mathrm{~d}}=4 \mathrm{~F}$ Capacitance of $\mathrm{I}^{\text {st }}$ half plate, $\mathrm{C}_{1}=\frac{\varepsilon_{0} \frac{\mathrm{A}}{2}}{\mathrm{~d}}=\frac{1}{2} \frac{\varepsilon_{0} \mathrm{~A}}{\mathrm{~d}}$ $\mathrm{C}_{1}=\frac{1}{2} \times 4=2 \mathrm{~F}$ Capacitance of $2^{\text {nd }}$ half plate, $\mathrm{C}_{2}=\frac{\mathrm{K} \varepsilon_{0} \frac{\mathrm{A}}{2}}{\mathrm{~d}}$ $=\left(\frac{\mathrm{K}}{2} \varepsilon_{0} \frac{\mathrm{A}}{\mathrm{d}}\right)=\frac{3}{2} \times 4=6 \mathrm{~F}$ According to question, $\mathrm{C}_{1}$ and $\mathrm{C}_{2}$ are in parallel- $\therefore \mathrm{C}=\mathrm{C}_{1}+\mathrm{C}_{2}$ $=2+6$ $=8 \mathrm{~F}$
[EAMCET-1993]
Capacitance
166031
A parallel plate condenser with oil between the plates (dielectric constant of oil $K=2$ ) has a capacitance $C$. It the oil is removed, the capacitance of the capacitor becomes
1 $\sqrt{2} \mathrm{C}$
2 $2 \mathrm{C}$
3 $\frac{\mathrm{C}}{\sqrt{2}}$
4 $\frac{\mathrm{C}}{2}$
Explanation:
: The Capacitance of a parallel plate capacitor with dielectric (oil) between its plates is $\mathrm{C}=\frac{\mathrm{K} \varepsilon_{0} \mathrm{~A}}{\mathrm{~d}}$ When dielectric (oil) is removed, so capacitance $\mathrm{C}_{0}=\frac{\varepsilon_{0} \mathrm{~A}}{\mathrm{~d}}$ Comparing equation (i) \& (ii), we get $\mathrm{C}=\mathrm{KC}_{0}$ $\mathrm{C}_{0}=\frac{\mathrm{C}}{\mathrm{K}}=\frac{\mathrm{C}}{2} \quad \therefore[\mathrm{K}=2]$
[EAMCET-1998]
Capacitance
166032
A parallel plate capacitor filled with a dielectric of relative permittivity 5 between its plates is charged to acquire an energy $E$ and isolated. If the dielectric is replaced by another of relative permittivity 2 , its energy becomes
1 $\mathrm{E}$
2 $0.4 \mathrm{E}$
3 $2.5 \mathrm{E}$
4 $6.25 \mathrm{E}$
Explanation:
: Energy stored in capacitor $(E)=\frac{q^{2}}{2 C}$ As charged is conserved so $\mathrm{E} \propto \frac{1}{\mathrm{C}}$ Let $\mathrm{C}_{0}$ be the capacitance of air filled capacitor When the dielectric material with dielectric constant $\mathrm{K}$ is filled between the places, the capacitance becomes $\mathrm{C}^{\prime}=\mathrm{K} \mathrm{C}_{0}$ $\frac{\mathrm{E}_{2}}{\mathrm{E}_{1}}=\frac{5 \mathrm{C}_{0}}{2 \mathrm{C}_{0}}=2.5$ $=\frac{\mathrm{E}_{2}}{\mathrm{E}}=2.5 \Rightarrow \mathrm{E}_{2}=2.5 \mathrm{E}$
NEET Test Series from KOTA - 10 Papers In MS WORD
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Capacitance
166028
Two capacitors of capacitance $C$ are connected in series If one of them is filled with dielectric substance $k$, what is the effective capacitance ?
1 $\frac{\mathrm{kC}}{(1+\mathrm{k})}$
2 $\mathrm{C}(\mathrm{k}+1)$
3 $\frac{2 \mathrm{kC}}{1+\mathrm{k}}$
4 none of these
Explanation:
: When plates of Capacitor are separated by a dielectric Medium of dielectric Constant k, its capacity $\mathrm{C}_{\mathrm{m}}=\frac{\mathrm{k} \varepsilon_{0} \mathrm{~A}}{\mathrm{~d}}=\mathrm{kC}_{0}=\mathrm{kC} \quad\left[\right.$ Where, $\left.\mathrm{C}_{0}=\mathrm{C}\right]$ Now, two capacitor $\mathrm{kC}$ and $\mathrm{C}$ are in Series, that effect Capacitance, $\frac{1}{\mathrm{C}}=\frac{1}{\mathrm{kC}}+\frac{1}{\mathrm{C}}$ $\Rightarrow \frac{1}{\mathrm{C}^{\prime}}=\frac{1+\mathrm{k}}{\mathrm{kC}}$ $\Rightarrow \mathrm{C}^{\prime}=\frac{\mathrm{kC}}{\mathrm{k}+1}$
[DCE-2007]
Capacitance
166030
In a parallel plate air capacitor of capacitance $4 \mathrm{~F}$ if the lower half of air space is filled with a material of dielectric constant 3 , its capacitance changes to
1 $\frac{4}{3} \mathrm{~F}$
2 $\frac{8}{3} \mathrm{~F}$
3 $8 \mathrm{~F}$
4 $12 \mathrm{~F}$
Explanation:
: Net capacitance, $\mathrm{C}=\mathrm{C}_{1}+\mathrm{C}_{2} \quad \ldots$ (i) Given, air capacitance $\mathrm{C}=\frac{\varepsilon_{0} \mathrm{~A}}{\mathrm{~d}}=4 \mathrm{~F}$ Capacitance of $\mathrm{I}^{\text {st }}$ half plate, $\mathrm{C}_{1}=\frac{\varepsilon_{0} \frac{\mathrm{A}}{2}}{\mathrm{~d}}=\frac{1}{2} \frac{\varepsilon_{0} \mathrm{~A}}{\mathrm{~d}}$ $\mathrm{C}_{1}=\frac{1}{2} \times 4=2 \mathrm{~F}$ Capacitance of $2^{\text {nd }}$ half plate, $\mathrm{C}_{2}=\frac{\mathrm{K} \varepsilon_{0} \frac{\mathrm{A}}{2}}{\mathrm{~d}}$ $=\left(\frac{\mathrm{K}}{2} \varepsilon_{0} \frac{\mathrm{A}}{\mathrm{d}}\right)=\frac{3}{2} \times 4=6 \mathrm{~F}$ According to question, $\mathrm{C}_{1}$ and $\mathrm{C}_{2}$ are in parallel- $\therefore \mathrm{C}=\mathrm{C}_{1}+\mathrm{C}_{2}$ $=2+6$ $=8 \mathrm{~F}$
[EAMCET-1993]
Capacitance
166031
A parallel plate condenser with oil between the plates (dielectric constant of oil $K=2$ ) has a capacitance $C$. It the oil is removed, the capacitance of the capacitor becomes
1 $\sqrt{2} \mathrm{C}$
2 $2 \mathrm{C}$
3 $\frac{\mathrm{C}}{\sqrt{2}}$
4 $\frac{\mathrm{C}}{2}$
Explanation:
: The Capacitance of a parallel plate capacitor with dielectric (oil) between its plates is $\mathrm{C}=\frac{\mathrm{K} \varepsilon_{0} \mathrm{~A}}{\mathrm{~d}}$ When dielectric (oil) is removed, so capacitance $\mathrm{C}_{0}=\frac{\varepsilon_{0} \mathrm{~A}}{\mathrm{~d}}$ Comparing equation (i) \& (ii), we get $\mathrm{C}=\mathrm{KC}_{0}$ $\mathrm{C}_{0}=\frac{\mathrm{C}}{\mathrm{K}}=\frac{\mathrm{C}}{2} \quad \therefore[\mathrm{K}=2]$
[EAMCET-1998]
Capacitance
166032
A parallel plate capacitor filled with a dielectric of relative permittivity 5 between its plates is charged to acquire an energy $E$ and isolated. If the dielectric is replaced by another of relative permittivity 2 , its energy becomes
1 $\mathrm{E}$
2 $0.4 \mathrm{E}$
3 $2.5 \mathrm{E}$
4 $6.25 \mathrm{E}$
Explanation:
: Energy stored in capacitor $(E)=\frac{q^{2}}{2 C}$ As charged is conserved so $\mathrm{E} \propto \frac{1}{\mathrm{C}}$ Let $\mathrm{C}_{0}$ be the capacitance of air filled capacitor When the dielectric material with dielectric constant $\mathrm{K}$ is filled between the places, the capacitance becomes $\mathrm{C}^{\prime}=\mathrm{K} \mathrm{C}_{0}$ $\frac{\mathrm{E}_{2}}{\mathrm{E}_{1}}=\frac{5 \mathrm{C}_{0}}{2 \mathrm{C}_{0}}=2.5$ $=\frac{\mathrm{E}_{2}}{\mathrm{E}}=2.5 \Rightarrow \mathrm{E}_{2}=2.5 \mathrm{E}$
166028
Two capacitors of capacitance $C$ are connected in series If one of them is filled with dielectric substance $k$, what is the effective capacitance ?
1 $\frac{\mathrm{kC}}{(1+\mathrm{k})}$
2 $\mathrm{C}(\mathrm{k}+1)$
3 $\frac{2 \mathrm{kC}}{1+\mathrm{k}}$
4 none of these
Explanation:
: When plates of Capacitor are separated by a dielectric Medium of dielectric Constant k, its capacity $\mathrm{C}_{\mathrm{m}}=\frac{\mathrm{k} \varepsilon_{0} \mathrm{~A}}{\mathrm{~d}}=\mathrm{kC}_{0}=\mathrm{kC} \quad\left[\right.$ Where, $\left.\mathrm{C}_{0}=\mathrm{C}\right]$ Now, two capacitor $\mathrm{kC}$ and $\mathrm{C}$ are in Series, that effect Capacitance, $\frac{1}{\mathrm{C}}=\frac{1}{\mathrm{kC}}+\frac{1}{\mathrm{C}}$ $\Rightarrow \frac{1}{\mathrm{C}^{\prime}}=\frac{1+\mathrm{k}}{\mathrm{kC}}$ $\Rightarrow \mathrm{C}^{\prime}=\frac{\mathrm{kC}}{\mathrm{k}+1}$
[DCE-2007]
Capacitance
166030
In a parallel plate air capacitor of capacitance $4 \mathrm{~F}$ if the lower half of air space is filled with a material of dielectric constant 3 , its capacitance changes to
1 $\frac{4}{3} \mathrm{~F}$
2 $\frac{8}{3} \mathrm{~F}$
3 $8 \mathrm{~F}$
4 $12 \mathrm{~F}$
Explanation:
: Net capacitance, $\mathrm{C}=\mathrm{C}_{1}+\mathrm{C}_{2} \quad \ldots$ (i) Given, air capacitance $\mathrm{C}=\frac{\varepsilon_{0} \mathrm{~A}}{\mathrm{~d}}=4 \mathrm{~F}$ Capacitance of $\mathrm{I}^{\text {st }}$ half plate, $\mathrm{C}_{1}=\frac{\varepsilon_{0} \frac{\mathrm{A}}{2}}{\mathrm{~d}}=\frac{1}{2} \frac{\varepsilon_{0} \mathrm{~A}}{\mathrm{~d}}$ $\mathrm{C}_{1}=\frac{1}{2} \times 4=2 \mathrm{~F}$ Capacitance of $2^{\text {nd }}$ half plate, $\mathrm{C}_{2}=\frac{\mathrm{K} \varepsilon_{0} \frac{\mathrm{A}}{2}}{\mathrm{~d}}$ $=\left(\frac{\mathrm{K}}{2} \varepsilon_{0} \frac{\mathrm{A}}{\mathrm{d}}\right)=\frac{3}{2} \times 4=6 \mathrm{~F}$ According to question, $\mathrm{C}_{1}$ and $\mathrm{C}_{2}$ are in parallel- $\therefore \mathrm{C}=\mathrm{C}_{1}+\mathrm{C}_{2}$ $=2+6$ $=8 \mathrm{~F}$
[EAMCET-1993]
Capacitance
166031
A parallel plate condenser with oil between the plates (dielectric constant of oil $K=2$ ) has a capacitance $C$. It the oil is removed, the capacitance of the capacitor becomes
1 $\sqrt{2} \mathrm{C}$
2 $2 \mathrm{C}$
3 $\frac{\mathrm{C}}{\sqrt{2}}$
4 $\frac{\mathrm{C}}{2}$
Explanation:
: The Capacitance of a parallel plate capacitor with dielectric (oil) between its plates is $\mathrm{C}=\frac{\mathrm{K} \varepsilon_{0} \mathrm{~A}}{\mathrm{~d}}$ When dielectric (oil) is removed, so capacitance $\mathrm{C}_{0}=\frac{\varepsilon_{0} \mathrm{~A}}{\mathrm{~d}}$ Comparing equation (i) \& (ii), we get $\mathrm{C}=\mathrm{KC}_{0}$ $\mathrm{C}_{0}=\frac{\mathrm{C}}{\mathrm{K}}=\frac{\mathrm{C}}{2} \quad \therefore[\mathrm{K}=2]$
[EAMCET-1998]
Capacitance
166032
A parallel plate capacitor filled with a dielectric of relative permittivity 5 between its plates is charged to acquire an energy $E$ and isolated. If the dielectric is replaced by another of relative permittivity 2 , its energy becomes
1 $\mathrm{E}$
2 $0.4 \mathrm{E}$
3 $2.5 \mathrm{E}$
4 $6.25 \mathrm{E}$
Explanation:
: Energy stored in capacitor $(E)=\frac{q^{2}}{2 C}$ As charged is conserved so $\mathrm{E} \propto \frac{1}{\mathrm{C}}$ Let $\mathrm{C}_{0}$ be the capacitance of air filled capacitor When the dielectric material with dielectric constant $\mathrm{K}$ is filled between the places, the capacitance becomes $\mathrm{C}^{\prime}=\mathrm{K} \mathrm{C}_{0}$ $\frac{\mathrm{E}_{2}}{\mathrm{E}_{1}}=\frac{5 \mathrm{C}_{0}}{2 \mathrm{C}_{0}}=2.5$ $=\frac{\mathrm{E}_{2}}{\mathrm{E}}=2.5 \Rightarrow \mathrm{E}_{2}=2.5 \mathrm{E}$
166028
Two capacitors of capacitance $C$ are connected in series If one of them is filled with dielectric substance $k$, what is the effective capacitance ?
1 $\frac{\mathrm{kC}}{(1+\mathrm{k})}$
2 $\mathrm{C}(\mathrm{k}+1)$
3 $\frac{2 \mathrm{kC}}{1+\mathrm{k}}$
4 none of these
Explanation:
: When plates of Capacitor are separated by a dielectric Medium of dielectric Constant k, its capacity $\mathrm{C}_{\mathrm{m}}=\frac{\mathrm{k} \varepsilon_{0} \mathrm{~A}}{\mathrm{~d}}=\mathrm{kC}_{0}=\mathrm{kC} \quad\left[\right.$ Where, $\left.\mathrm{C}_{0}=\mathrm{C}\right]$ Now, two capacitor $\mathrm{kC}$ and $\mathrm{C}$ are in Series, that effect Capacitance, $\frac{1}{\mathrm{C}}=\frac{1}{\mathrm{kC}}+\frac{1}{\mathrm{C}}$ $\Rightarrow \frac{1}{\mathrm{C}^{\prime}}=\frac{1+\mathrm{k}}{\mathrm{kC}}$ $\Rightarrow \mathrm{C}^{\prime}=\frac{\mathrm{kC}}{\mathrm{k}+1}$
[DCE-2007]
Capacitance
166030
In a parallel plate air capacitor of capacitance $4 \mathrm{~F}$ if the lower half of air space is filled with a material of dielectric constant 3 , its capacitance changes to
1 $\frac{4}{3} \mathrm{~F}$
2 $\frac{8}{3} \mathrm{~F}$
3 $8 \mathrm{~F}$
4 $12 \mathrm{~F}$
Explanation:
: Net capacitance, $\mathrm{C}=\mathrm{C}_{1}+\mathrm{C}_{2} \quad \ldots$ (i) Given, air capacitance $\mathrm{C}=\frac{\varepsilon_{0} \mathrm{~A}}{\mathrm{~d}}=4 \mathrm{~F}$ Capacitance of $\mathrm{I}^{\text {st }}$ half plate, $\mathrm{C}_{1}=\frac{\varepsilon_{0} \frac{\mathrm{A}}{2}}{\mathrm{~d}}=\frac{1}{2} \frac{\varepsilon_{0} \mathrm{~A}}{\mathrm{~d}}$ $\mathrm{C}_{1}=\frac{1}{2} \times 4=2 \mathrm{~F}$ Capacitance of $2^{\text {nd }}$ half plate, $\mathrm{C}_{2}=\frac{\mathrm{K} \varepsilon_{0} \frac{\mathrm{A}}{2}}{\mathrm{~d}}$ $=\left(\frac{\mathrm{K}}{2} \varepsilon_{0} \frac{\mathrm{A}}{\mathrm{d}}\right)=\frac{3}{2} \times 4=6 \mathrm{~F}$ According to question, $\mathrm{C}_{1}$ and $\mathrm{C}_{2}$ are in parallel- $\therefore \mathrm{C}=\mathrm{C}_{1}+\mathrm{C}_{2}$ $=2+6$ $=8 \mathrm{~F}$
[EAMCET-1993]
Capacitance
166031
A parallel plate condenser with oil between the plates (dielectric constant of oil $K=2$ ) has a capacitance $C$. It the oil is removed, the capacitance of the capacitor becomes
1 $\sqrt{2} \mathrm{C}$
2 $2 \mathrm{C}$
3 $\frac{\mathrm{C}}{\sqrt{2}}$
4 $\frac{\mathrm{C}}{2}$
Explanation:
: The Capacitance of a parallel plate capacitor with dielectric (oil) between its plates is $\mathrm{C}=\frac{\mathrm{K} \varepsilon_{0} \mathrm{~A}}{\mathrm{~d}}$ When dielectric (oil) is removed, so capacitance $\mathrm{C}_{0}=\frac{\varepsilon_{0} \mathrm{~A}}{\mathrm{~d}}$ Comparing equation (i) \& (ii), we get $\mathrm{C}=\mathrm{KC}_{0}$ $\mathrm{C}_{0}=\frac{\mathrm{C}}{\mathrm{K}}=\frac{\mathrm{C}}{2} \quad \therefore[\mathrm{K}=2]$
[EAMCET-1998]
Capacitance
166032
A parallel plate capacitor filled with a dielectric of relative permittivity 5 between its plates is charged to acquire an energy $E$ and isolated. If the dielectric is replaced by another of relative permittivity 2 , its energy becomes
1 $\mathrm{E}$
2 $0.4 \mathrm{E}$
3 $2.5 \mathrm{E}$
4 $6.25 \mathrm{E}$
Explanation:
: Energy stored in capacitor $(E)=\frac{q^{2}}{2 C}$ As charged is conserved so $\mathrm{E} \propto \frac{1}{\mathrm{C}}$ Let $\mathrm{C}_{0}$ be the capacitance of air filled capacitor When the dielectric material with dielectric constant $\mathrm{K}$ is filled between the places, the capacitance becomes $\mathrm{C}^{\prime}=\mathrm{K} \mathrm{C}_{0}$ $\frac{\mathrm{E}_{2}}{\mathrm{E}_{1}}=\frac{5 \mathrm{C}_{0}}{2 \mathrm{C}_{0}}=2.5$ $=\frac{\mathrm{E}_{2}}{\mathrm{E}}=2.5 \Rightarrow \mathrm{E}_{2}=2.5 \mathrm{E}$