166002
A parallel plate capacitor has area $2 \mathrm{~m}^{2}$ separated by 3 dielectric slabs. Their relative permittivity is $2,3,6$ and thickness is $0.4 \mathrm{~mm}$, $0.6 \mathrm{~mm}, 1.2 \mathrm{~mm}$ respectively. The capacitance is
166003
A capacitor is connected to a battery of voltage V. Now a dielectric slab of dielectric constant $K$ is completely inserted between the plates, then the final charge on the capacitor will be:
: Given that, Voltage $=\mathrm{V}$ Dielectric constant $=\mathrm{K}$ We know that, $\mathrm{C}=\frac{\mathrm{K} \in_{0} \mathrm{~A}}{\mathrm{~d}}$ Then, the final charge on the capacitor $\mathrm{Q}=\mathrm{C} \times \mathrm{V}$ Putting the value of $\mathrm{C}$ from equation (i) $\mathrm{Q}=\left(\frac{\mathrm{K} \in_{0} \mathrm{~A}}{\mathrm{~d}}\right) \mathrm{V}$
[AIIMS-26.05.2019(E)]**# Shift-2]
Capacitance
166004
A capacitor of capacitance $9 \mathrm{nF}$ having dielectric slab of $\varepsilon_{\mathrm{r}}=2.4$, dielectric strength 20 $\mathrm{MV} / \mathrm{m}$ and P.D. $=20 \mathrm{~V}$. Calculate the area of plates.
166005
A capacitor of capacitance $15 \mathrm{nF}$ have dielectric slab of dielectric constant 2.5 , electric field strength $30 \mathrm{MV} / \mathrm{m}$ and potential difference 30 volt. Calculate the area of plate.
166006
When capacitor is fully charged, find current drawn from the cell.
1 $2 \mathrm{~mA}$
2 $1 \mathrm{~mA}$
3 $3 \mathrm{~mA}$
4 $9 \mathrm{~mA}$
Explanation:
: When capacitor is fully charged then it becomes open circuit hence no current flow. so the equivalent circuit is- $\mathrm{R}_{\mathrm{eq}}=\mathrm{R}_{1}+\mathrm{R}_{2}=1 \mathrm{~K} \Omega+2 \mathrm{k} \Omega=3 \mathrm{k} \Omega$ Hence, current draw from the cell- $\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}_{\mathrm{eq}}}=\frac{9}{3 \times 10^{3}}=3 \times 10^{-3}=3 \mathrm{~mA}$
166002
A parallel plate capacitor has area $2 \mathrm{~m}^{2}$ separated by 3 dielectric slabs. Their relative permittivity is $2,3,6$ and thickness is $0.4 \mathrm{~mm}$, $0.6 \mathrm{~mm}, 1.2 \mathrm{~mm}$ respectively. The capacitance is
166003
A capacitor is connected to a battery of voltage V. Now a dielectric slab of dielectric constant $K$ is completely inserted between the plates, then the final charge on the capacitor will be:
: Given that, Voltage $=\mathrm{V}$ Dielectric constant $=\mathrm{K}$ We know that, $\mathrm{C}=\frac{\mathrm{K} \in_{0} \mathrm{~A}}{\mathrm{~d}}$ Then, the final charge on the capacitor $\mathrm{Q}=\mathrm{C} \times \mathrm{V}$ Putting the value of $\mathrm{C}$ from equation (i) $\mathrm{Q}=\left(\frac{\mathrm{K} \in_{0} \mathrm{~A}}{\mathrm{~d}}\right) \mathrm{V}$
[AIIMS-26.05.2019(E)]**# Shift-2]
Capacitance
166004
A capacitor of capacitance $9 \mathrm{nF}$ having dielectric slab of $\varepsilon_{\mathrm{r}}=2.4$, dielectric strength 20 $\mathrm{MV} / \mathrm{m}$ and P.D. $=20 \mathrm{~V}$. Calculate the area of plates.
166005
A capacitor of capacitance $15 \mathrm{nF}$ have dielectric slab of dielectric constant 2.5 , electric field strength $30 \mathrm{MV} / \mathrm{m}$ and potential difference 30 volt. Calculate the area of plate.
166006
When capacitor is fully charged, find current drawn from the cell.
1 $2 \mathrm{~mA}$
2 $1 \mathrm{~mA}$
3 $3 \mathrm{~mA}$
4 $9 \mathrm{~mA}$
Explanation:
: When capacitor is fully charged then it becomes open circuit hence no current flow. so the equivalent circuit is- $\mathrm{R}_{\mathrm{eq}}=\mathrm{R}_{1}+\mathrm{R}_{2}=1 \mathrm{~K} \Omega+2 \mathrm{k} \Omega=3 \mathrm{k} \Omega$ Hence, current draw from the cell- $\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}_{\mathrm{eq}}}=\frac{9}{3 \times 10^{3}}=3 \times 10^{-3}=3 \mathrm{~mA}$
166002
A parallel plate capacitor has area $2 \mathrm{~m}^{2}$ separated by 3 dielectric slabs. Their relative permittivity is $2,3,6$ and thickness is $0.4 \mathrm{~mm}$, $0.6 \mathrm{~mm}, 1.2 \mathrm{~mm}$ respectively. The capacitance is
166003
A capacitor is connected to a battery of voltage V. Now a dielectric slab of dielectric constant $K$ is completely inserted between the plates, then the final charge on the capacitor will be:
: Given that, Voltage $=\mathrm{V}$ Dielectric constant $=\mathrm{K}$ We know that, $\mathrm{C}=\frac{\mathrm{K} \in_{0} \mathrm{~A}}{\mathrm{~d}}$ Then, the final charge on the capacitor $\mathrm{Q}=\mathrm{C} \times \mathrm{V}$ Putting the value of $\mathrm{C}$ from equation (i) $\mathrm{Q}=\left(\frac{\mathrm{K} \in_{0} \mathrm{~A}}{\mathrm{~d}}\right) \mathrm{V}$
[AIIMS-26.05.2019(E)]**# Shift-2]
Capacitance
166004
A capacitor of capacitance $9 \mathrm{nF}$ having dielectric slab of $\varepsilon_{\mathrm{r}}=2.4$, dielectric strength 20 $\mathrm{MV} / \mathrm{m}$ and P.D. $=20 \mathrm{~V}$. Calculate the area of plates.
166005
A capacitor of capacitance $15 \mathrm{nF}$ have dielectric slab of dielectric constant 2.5 , electric field strength $30 \mathrm{MV} / \mathrm{m}$ and potential difference 30 volt. Calculate the area of plate.
166006
When capacitor is fully charged, find current drawn from the cell.
1 $2 \mathrm{~mA}$
2 $1 \mathrm{~mA}$
3 $3 \mathrm{~mA}$
4 $9 \mathrm{~mA}$
Explanation:
: When capacitor is fully charged then it becomes open circuit hence no current flow. so the equivalent circuit is- $\mathrm{R}_{\mathrm{eq}}=\mathrm{R}_{1}+\mathrm{R}_{2}=1 \mathrm{~K} \Omega+2 \mathrm{k} \Omega=3 \mathrm{k} \Omega$ Hence, current draw from the cell- $\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}_{\mathrm{eq}}}=\frac{9}{3 \times 10^{3}}=3 \times 10^{-3}=3 \mathrm{~mA}$
166002
A parallel plate capacitor has area $2 \mathrm{~m}^{2}$ separated by 3 dielectric slabs. Their relative permittivity is $2,3,6$ and thickness is $0.4 \mathrm{~mm}$, $0.6 \mathrm{~mm}, 1.2 \mathrm{~mm}$ respectively. The capacitance is
166003
A capacitor is connected to a battery of voltage V. Now a dielectric slab of dielectric constant $K$ is completely inserted between the plates, then the final charge on the capacitor will be:
: Given that, Voltage $=\mathrm{V}$ Dielectric constant $=\mathrm{K}$ We know that, $\mathrm{C}=\frac{\mathrm{K} \in_{0} \mathrm{~A}}{\mathrm{~d}}$ Then, the final charge on the capacitor $\mathrm{Q}=\mathrm{C} \times \mathrm{V}$ Putting the value of $\mathrm{C}$ from equation (i) $\mathrm{Q}=\left(\frac{\mathrm{K} \in_{0} \mathrm{~A}}{\mathrm{~d}}\right) \mathrm{V}$
[AIIMS-26.05.2019(E)]**# Shift-2]
Capacitance
166004
A capacitor of capacitance $9 \mathrm{nF}$ having dielectric slab of $\varepsilon_{\mathrm{r}}=2.4$, dielectric strength 20 $\mathrm{MV} / \mathrm{m}$ and P.D. $=20 \mathrm{~V}$. Calculate the area of plates.
166005
A capacitor of capacitance $15 \mathrm{nF}$ have dielectric slab of dielectric constant 2.5 , electric field strength $30 \mathrm{MV} / \mathrm{m}$ and potential difference 30 volt. Calculate the area of plate.
166006
When capacitor is fully charged, find current drawn from the cell.
1 $2 \mathrm{~mA}$
2 $1 \mathrm{~mA}$
3 $3 \mathrm{~mA}$
4 $9 \mathrm{~mA}$
Explanation:
: When capacitor is fully charged then it becomes open circuit hence no current flow. so the equivalent circuit is- $\mathrm{R}_{\mathrm{eq}}=\mathrm{R}_{1}+\mathrm{R}_{2}=1 \mathrm{~K} \Omega+2 \mathrm{k} \Omega=3 \mathrm{k} \Omega$ Hence, current draw from the cell- $\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}_{\mathrm{eq}}}=\frac{9}{3 \times 10^{3}}=3 \times 10^{-3}=3 \mathrm{~mA}$
166002
A parallel plate capacitor has area $2 \mathrm{~m}^{2}$ separated by 3 dielectric slabs. Their relative permittivity is $2,3,6$ and thickness is $0.4 \mathrm{~mm}$, $0.6 \mathrm{~mm}, 1.2 \mathrm{~mm}$ respectively. The capacitance is
166003
A capacitor is connected to a battery of voltage V. Now a dielectric slab of dielectric constant $K$ is completely inserted between the plates, then the final charge on the capacitor will be:
: Given that, Voltage $=\mathrm{V}$ Dielectric constant $=\mathrm{K}$ We know that, $\mathrm{C}=\frac{\mathrm{K} \in_{0} \mathrm{~A}}{\mathrm{~d}}$ Then, the final charge on the capacitor $\mathrm{Q}=\mathrm{C} \times \mathrm{V}$ Putting the value of $\mathrm{C}$ from equation (i) $\mathrm{Q}=\left(\frac{\mathrm{K} \in_{0} \mathrm{~A}}{\mathrm{~d}}\right) \mathrm{V}$
[AIIMS-26.05.2019(E)]**# Shift-2]
Capacitance
166004
A capacitor of capacitance $9 \mathrm{nF}$ having dielectric slab of $\varepsilon_{\mathrm{r}}=2.4$, dielectric strength 20 $\mathrm{MV} / \mathrm{m}$ and P.D. $=20 \mathrm{~V}$. Calculate the area of plates.
166005
A capacitor of capacitance $15 \mathrm{nF}$ have dielectric slab of dielectric constant 2.5 , electric field strength $30 \mathrm{MV} / \mathrm{m}$ and potential difference 30 volt. Calculate the area of plate.
166006
When capacitor is fully charged, find current drawn from the cell.
1 $2 \mathrm{~mA}$
2 $1 \mathrm{~mA}$
3 $3 \mathrm{~mA}$
4 $9 \mathrm{~mA}$
Explanation:
: When capacitor is fully charged then it becomes open circuit hence no current flow. so the equivalent circuit is- $\mathrm{R}_{\mathrm{eq}}=\mathrm{R}_{1}+\mathrm{R}_{2}=1 \mathrm{~K} \Omega+2 \mathrm{k} \Omega=3 \mathrm{k} \Omega$ Hence, current draw from the cell- $\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}_{\mathrm{eq}}}=\frac{9}{3 \times 10^{3}}=3 \times 10^{-3}=3 \mathrm{~mA}$
