166011
A condenser of capacity $C_{1}$ is charged to potential $V_{1}$ and then disconnected. Uncharged capacitor of capacity $\mathrm{C}_{2}$ is connected in parallel with $C_{1}$. The resultant potential $V_{2}$ is
: Charge on first capacitor $\left(\mathrm{q}_{1}\right)=\mathrm{C}_{1} \mathrm{~V}_{1}$ Charge on second capacitor $\left(\mathrm{q}_{2}\right)=0$ When they are connected in parallel then total charge $\mathrm{q}=\mathrm{q}_{1}+\mathrm{q}_{2} \quad\left(\therefore \mathrm{q}=\mathrm{C}_{1} \mathrm{~V}_{1}\right)$ And capacitance $(C)=\mathrm{C}_{1}+\mathrm{C}_{2}$ Let, $\mathrm{V}^{\prime}$ be the common potential difference across each capacitor then $\mathrm{q}=\mathrm{CV}^{\prime}$ $\therefore \quad \mathrm{V}^{\prime}=\frac{\mathrm{q}}{\mathrm{C}}=\frac{\mathrm{C}_{1} \mathrm{~V}_{1}}{\mathrm{C}_{1}+\mathrm{C}_{2}}$
[MHT-CET 2020]
Capacitance
166012
The capacity of a parallel plate air capacitor is $2 \mu \mathrm{F}$ and voltage between the plates is changing at the rate of $3 \mathrm{~V} / \mathrm{s}$. The displacement current in the capacitor is
1 $2 \mu \mathrm{A}$
2 $3 \mu \mathrm{A}$
3 $5 \mu \mathrm{A}$
4 $6 \mu \mathrm{A}$
Explanation:
Given, Capacitance, $\mathrm{C}=2 \mu \mathrm{F}, \frac{\mathrm{dV}}{\mathrm{dt}}=3 \mathrm{~V} / \mathrm{s}$ Charge across the plate of capacitor $\mathrm{q}=\mathrm{CV}$ Differentiating with respect to time $\frac{\mathrm{dq}}{\mathrm{dt}}=\mathrm{C} \frac{\mathrm{dV}}{\mathrm{dt}}$ Displacement current $\mathrm{i}=\mathrm{C} \frac{\mathrm{dV}}{\mathrm{dt}}$ $=2 \mu \mathrm{F} \times 3 \mathrm{~V} / \mathrm{s}=6 \mu \mathrm{A}$
[MHT-CET 2015]
Capacitance
166013
The capacitance of a parallel plate capacitor with air as medium is $3 \mu \mathrm{F}$. As a dielectric is introduced between the plates, the capacitance becomes $15 \mu \mathrm{F}$. The permittivity of the medium in $\mathrm{C}^{2} \mathbf{N}^{-1} \mathrm{~m}^{-2}$ is
1 $8.15 \times 10^{-11}$
2 $0.44 \times 10^{-10}$
3 $15.2 \times 10^{12}$
4 $1.6 \times 10^{-14}$
Explanation:
: The capacitance of the capacitor is written as $\mathrm{C}=\frac{\varepsilon_{0} \mathrm{~A}}{\mathrm{~d}}$ Capacitance of a same parallel plate capacitor with the introduction of a dielectric medium- $\mathrm{C}^{\prime}=\frac{\mathrm{K} \varepsilon_{0} \mathrm{~A}}{\mathrm{~d}}$ Where $\mathrm{K}$ is the dielectric constant of a medium So, $\quad \frac{\mathrm{C}^{\prime}}{\mathrm{C}}=\mathrm{K} \Rightarrow \mathrm{K}=\frac{15}{3}=5$ Now, $\mathrm{K}=\frac{\varepsilon}{\varepsilon_{\mathrm{o}}}$ $\varepsilon=\mathrm{K} \varepsilon_{\mathrm{o}} =5 \times 8.854 \times 10^{-12}$ $=0.44 \times 10^{-10} \mathrm{C}^{2} \mathrm{~N}^{-1} \mathrm{~m}^{-2}$
[VITEEE-2015]
Capacitance
166014
The capacity of a capacitor is $4 \times 10^{-6} \mathrm{~F}$ and its potential is $100 \mathrm{~V}$. The energy released on discharging it fully will be
1 $0.02 \mathrm{~J}$
2 $0.04 \mathrm{~J}$
3 $0.025 \mathrm{~J}$
4 $0.05 \mathrm{~J}$
Explanation:
: Given that, Capacitance, $\mathrm{C}=4 \times 10^{-6} \mathrm{~F}$ Potential $=100 \mathrm{~V}$ Energy released on discharging the capacitor- $\mathrm{E} =\frac{1}{2} \mathrm{CV}^{2}$ $=\frac{1}{2} \times 4 \times 10^{-6} \times(100)^{2}$ $=0.02 \text { Jule }$
166011
A condenser of capacity $C_{1}$ is charged to potential $V_{1}$ and then disconnected. Uncharged capacitor of capacity $\mathrm{C}_{2}$ is connected in parallel with $C_{1}$. The resultant potential $V_{2}$ is
: Charge on first capacitor $\left(\mathrm{q}_{1}\right)=\mathrm{C}_{1} \mathrm{~V}_{1}$ Charge on second capacitor $\left(\mathrm{q}_{2}\right)=0$ When they are connected in parallel then total charge $\mathrm{q}=\mathrm{q}_{1}+\mathrm{q}_{2} \quad\left(\therefore \mathrm{q}=\mathrm{C}_{1} \mathrm{~V}_{1}\right)$ And capacitance $(C)=\mathrm{C}_{1}+\mathrm{C}_{2}$ Let, $\mathrm{V}^{\prime}$ be the common potential difference across each capacitor then $\mathrm{q}=\mathrm{CV}^{\prime}$ $\therefore \quad \mathrm{V}^{\prime}=\frac{\mathrm{q}}{\mathrm{C}}=\frac{\mathrm{C}_{1} \mathrm{~V}_{1}}{\mathrm{C}_{1}+\mathrm{C}_{2}}$
[MHT-CET 2020]
Capacitance
166012
The capacity of a parallel plate air capacitor is $2 \mu \mathrm{F}$ and voltage between the plates is changing at the rate of $3 \mathrm{~V} / \mathrm{s}$. The displacement current in the capacitor is
1 $2 \mu \mathrm{A}$
2 $3 \mu \mathrm{A}$
3 $5 \mu \mathrm{A}$
4 $6 \mu \mathrm{A}$
Explanation:
Given, Capacitance, $\mathrm{C}=2 \mu \mathrm{F}, \frac{\mathrm{dV}}{\mathrm{dt}}=3 \mathrm{~V} / \mathrm{s}$ Charge across the plate of capacitor $\mathrm{q}=\mathrm{CV}$ Differentiating with respect to time $\frac{\mathrm{dq}}{\mathrm{dt}}=\mathrm{C} \frac{\mathrm{dV}}{\mathrm{dt}}$ Displacement current $\mathrm{i}=\mathrm{C} \frac{\mathrm{dV}}{\mathrm{dt}}$ $=2 \mu \mathrm{F} \times 3 \mathrm{~V} / \mathrm{s}=6 \mu \mathrm{A}$
[MHT-CET 2015]
Capacitance
166013
The capacitance of a parallel plate capacitor with air as medium is $3 \mu \mathrm{F}$. As a dielectric is introduced between the plates, the capacitance becomes $15 \mu \mathrm{F}$. The permittivity of the medium in $\mathrm{C}^{2} \mathbf{N}^{-1} \mathrm{~m}^{-2}$ is
1 $8.15 \times 10^{-11}$
2 $0.44 \times 10^{-10}$
3 $15.2 \times 10^{12}$
4 $1.6 \times 10^{-14}$
Explanation:
: The capacitance of the capacitor is written as $\mathrm{C}=\frac{\varepsilon_{0} \mathrm{~A}}{\mathrm{~d}}$ Capacitance of a same parallel plate capacitor with the introduction of a dielectric medium- $\mathrm{C}^{\prime}=\frac{\mathrm{K} \varepsilon_{0} \mathrm{~A}}{\mathrm{~d}}$ Where $\mathrm{K}$ is the dielectric constant of a medium So, $\quad \frac{\mathrm{C}^{\prime}}{\mathrm{C}}=\mathrm{K} \Rightarrow \mathrm{K}=\frac{15}{3}=5$ Now, $\mathrm{K}=\frac{\varepsilon}{\varepsilon_{\mathrm{o}}}$ $\varepsilon=\mathrm{K} \varepsilon_{\mathrm{o}} =5 \times 8.854 \times 10^{-12}$ $=0.44 \times 10^{-10} \mathrm{C}^{2} \mathrm{~N}^{-1} \mathrm{~m}^{-2}$
[VITEEE-2015]
Capacitance
166014
The capacity of a capacitor is $4 \times 10^{-6} \mathrm{~F}$ and its potential is $100 \mathrm{~V}$. The energy released on discharging it fully will be
1 $0.02 \mathrm{~J}$
2 $0.04 \mathrm{~J}$
3 $0.025 \mathrm{~J}$
4 $0.05 \mathrm{~J}$
Explanation:
: Given that, Capacitance, $\mathrm{C}=4 \times 10^{-6} \mathrm{~F}$ Potential $=100 \mathrm{~V}$ Energy released on discharging the capacitor- $\mathrm{E} =\frac{1}{2} \mathrm{CV}^{2}$ $=\frac{1}{2} \times 4 \times 10^{-6} \times(100)^{2}$ $=0.02 \text { Jule }$
166011
A condenser of capacity $C_{1}$ is charged to potential $V_{1}$ and then disconnected. Uncharged capacitor of capacity $\mathrm{C}_{2}$ is connected in parallel with $C_{1}$. The resultant potential $V_{2}$ is
: Charge on first capacitor $\left(\mathrm{q}_{1}\right)=\mathrm{C}_{1} \mathrm{~V}_{1}$ Charge on second capacitor $\left(\mathrm{q}_{2}\right)=0$ When they are connected in parallel then total charge $\mathrm{q}=\mathrm{q}_{1}+\mathrm{q}_{2} \quad\left(\therefore \mathrm{q}=\mathrm{C}_{1} \mathrm{~V}_{1}\right)$ And capacitance $(C)=\mathrm{C}_{1}+\mathrm{C}_{2}$ Let, $\mathrm{V}^{\prime}$ be the common potential difference across each capacitor then $\mathrm{q}=\mathrm{CV}^{\prime}$ $\therefore \quad \mathrm{V}^{\prime}=\frac{\mathrm{q}}{\mathrm{C}}=\frac{\mathrm{C}_{1} \mathrm{~V}_{1}}{\mathrm{C}_{1}+\mathrm{C}_{2}}$
[MHT-CET 2020]
Capacitance
166012
The capacity of a parallel plate air capacitor is $2 \mu \mathrm{F}$ and voltage between the plates is changing at the rate of $3 \mathrm{~V} / \mathrm{s}$. The displacement current in the capacitor is
1 $2 \mu \mathrm{A}$
2 $3 \mu \mathrm{A}$
3 $5 \mu \mathrm{A}$
4 $6 \mu \mathrm{A}$
Explanation:
Given, Capacitance, $\mathrm{C}=2 \mu \mathrm{F}, \frac{\mathrm{dV}}{\mathrm{dt}}=3 \mathrm{~V} / \mathrm{s}$ Charge across the plate of capacitor $\mathrm{q}=\mathrm{CV}$ Differentiating with respect to time $\frac{\mathrm{dq}}{\mathrm{dt}}=\mathrm{C} \frac{\mathrm{dV}}{\mathrm{dt}}$ Displacement current $\mathrm{i}=\mathrm{C} \frac{\mathrm{dV}}{\mathrm{dt}}$ $=2 \mu \mathrm{F} \times 3 \mathrm{~V} / \mathrm{s}=6 \mu \mathrm{A}$
[MHT-CET 2015]
Capacitance
166013
The capacitance of a parallel plate capacitor with air as medium is $3 \mu \mathrm{F}$. As a dielectric is introduced between the plates, the capacitance becomes $15 \mu \mathrm{F}$. The permittivity of the medium in $\mathrm{C}^{2} \mathbf{N}^{-1} \mathrm{~m}^{-2}$ is
1 $8.15 \times 10^{-11}$
2 $0.44 \times 10^{-10}$
3 $15.2 \times 10^{12}$
4 $1.6 \times 10^{-14}$
Explanation:
: The capacitance of the capacitor is written as $\mathrm{C}=\frac{\varepsilon_{0} \mathrm{~A}}{\mathrm{~d}}$ Capacitance of a same parallel plate capacitor with the introduction of a dielectric medium- $\mathrm{C}^{\prime}=\frac{\mathrm{K} \varepsilon_{0} \mathrm{~A}}{\mathrm{~d}}$ Where $\mathrm{K}$ is the dielectric constant of a medium So, $\quad \frac{\mathrm{C}^{\prime}}{\mathrm{C}}=\mathrm{K} \Rightarrow \mathrm{K}=\frac{15}{3}=5$ Now, $\mathrm{K}=\frac{\varepsilon}{\varepsilon_{\mathrm{o}}}$ $\varepsilon=\mathrm{K} \varepsilon_{\mathrm{o}} =5 \times 8.854 \times 10^{-12}$ $=0.44 \times 10^{-10} \mathrm{C}^{2} \mathrm{~N}^{-1} \mathrm{~m}^{-2}$
[VITEEE-2015]
Capacitance
166014
The capacity of a capacitor is $4 \times 10^{-6} \mathrm{~F}$ and its potential is $100 \mathrm{~V}$. The energy released on discharging it fully will be
1 $0.02 \mathrm{~J}$
2 $0.04 \mathrm{~J}$
3 $0.025 \mathrm{~J}$
4 $0.05 \mathrm{~J}$
Explanation:
: Given that, Capacitance, $\mathrm{C}=4 \times 10^{-6} \mathrm{~F}$ Potential $=100 \mathrm{~V}$ Energy released on discharging the capacitor- $\mathrm{E} =\frac{1}{2} \mathrm{CV}^{2}$ $=\frac{1}{2} \times 4 \times 10^{-6} \times(100)^{2}$ $=0.02 \text { Jule }$
166011
A condenser of capacity $C_{1}$ is charged to potential $V_{1}$ and then disconnected. Uncharged capacitor of capacity $\mathrm{C}_{2}$ is connected in parallel with $C_{1}$. The resultant potential $V_{2}$ is
: Charge on first capacitor $\left(\mathrm{q}_{1}\right)=\mathrm{C}_{1} \mathrm{~V}_{1}$ Charge on second capacitor $\left(\mathrm{q}_{2}\right)=0$ When they are connected in parallel then total charge $\mathrm{q}=\mathrm{q}_{1}+\mathrm{q}_{2} \quad\left(\therefore \mathrm{q}=\mathrm{C}_{1} \mathrm{~V}_{1}\right)$ And capacitance $(C)=\mathrm{C}_{1}+\mathrm{C}_{2}$ Let, $\mathrm{V}^{\prime}$ be the common potential difference across each capacitor then $\mathrm{q}=\mathrm{CV}^{\prime}$ $\therefore \quad \mathrm{V}^{\prime}=\frac{\mathrm{q}}{\mathrm{C}}=\frac{\mathrm{C}_{1} \mathrm{~V}_{1}}{\mathrm{C}_{1}+\mathrm{C}_{2}}$
[MHT-CET 2020]
Capacitance
166012
The capacity of a parallel plate air capacitor is $2 \mu \mathrm{F}$ and voltage between the plates is changing at the rate of $3 \mathrm{~V} / \mathrm{s}$. The displacement current in the capacitor is
1 $2 \mu \mathrm{A}$
2 $3 \mu \mathrm{A}$
3 $5 \mu \mathrm{A}$
4 $6 \mu \mathrm{A}$
Explanation:
Given, Capacitance, $\mathrm{C}=2 \mu \mathrm{F}, \frac{\mathrm{dV}}{\mathrm{dt}}=3 \mathrm{~V} / \mathrm{s}$ Charge across the plate of capacitor $\mathrm{q}=\mathrm{CV}$ Differentiating with respect to time $\frac{\mathrm{dq}}{\mathrm{dt}}=\mathrm{C} \frac{\mathrm{dV}}{\mathrm{dt}}$ Displacement current $\mathrm{i}=\mathrm{C} \frac{\mathrm{dV}}{\mathrm{dt}}$ $=2 \mu \mathrm{F} \times 3 \mathrm{~V} / \mathrm{s}=6 \mu \mathrm{A}$
[MHT-CET 2015]
Capacitance
166013
The capacitance of a parallel plate capacitor with air as medium is $3 \mu \mathrm{F}$. As a dielectric is introduced between the plates, the capacitance becomes $15 \mu \mathrm{F}$. The permittivity of the medium in $\mathrm{C}^{2} \mathbf{N}^{-1} \mathrm{~m}^{-2}$ is
1 $8.15 \times 10^{-11}$
2 $0.44 \times 10^{-10}$
3 $15.2 \times 10^{12}$
4 $1.6 \times 10^{-14}$
Explanation:
: The capacitance of the capacitor is written as $\mathrm{C}=\frac{\varepsilon_{0} \mathrm{~A}}{\mathrm{~d}}$ Capacitance of a same parallel plate capacitor with the introduction of a dielectric medium- $\mathrm{C}^{\prime}=\frac{\mathrm{K} \varepsilon_{0} \mathrm{~A}}{\mathrm{~d}}$ Where $\mathrm{K}$ is the dielectric constant of a medium So, $\quad \frac{\mathrm{C}^{\prime}}{\mathrm{C}}=\mathrm{K} \Rightarrow \mathrm{K}=\frac{15}{3}=5$ Now, $\mathrm{K}=\frac{\varepsilon}{\varepsilon_{\mathrm{o}}}$ $\varepsilon=\mathrm{K} \varepsilon_{\mathrm{o}} =5 \times 8.854 \times 10^{-12}$ $=0.44 \times 10^{-10} \mathrm{C}^{2} \mathrm{~N}^{-1} \mathrm{~m}^{-2}$
[VITEEE-2015]
Capacitance
166014
The capacity of a capacitor is $4 \times 10^{-6} \mathrm{~F}$ and its potential is $100 \mathrm{~V}$. The energy released on discharging it fully will be
1 $0.02 \mathrm{~J}$
2 $0.04 \mathrm{~J}$
3 $0.025 \mathrm{~J}$
4 $0.05 \mathrm{~J}$
Explanation:
: Given that, Capacitance, $\mathrm{C}=4 \times 10^{-6} \mathrm{~F}$ Potential $=100 \mathrm{~V}$ Energy released on discharging the capacitor- $\mathrm{E} =\frac{1}{2} \mathrm{CV}^{2}$ $=\frac{1}{2} \times 4 \times 10^{-6} \times(100)^{2}$ $=0.02 \text { Jule }$
