166002
A parallel plate capacitor has area separated by 3 dielectric slabs. Their relative permittivity is and thickness is , respectively. The capacitance is
1 Farad
2 Farad
3 Farad
4 Farad
Explanation:
: Given, Area (A) And Then, equivalent capacitance,
[COMEDK 2015]
Capacitance
166003
A capacitor is connected to a battery of voltage V. Now a dielectric slab of dielectric constant is completely inserted between the plates, then the final charge on the capacitor will be:
1
2
3
4 zero
Explanation:
: Given that, Voltage Dielectric constant We know that, Then, the final charge on the capacitor Putting the value of from equation (i)
[AIIMS-26.05.2019(E)]**# Shift-2]
Capacitance
166004
A capacitor of capacitance having dielectric slab of , dielectric strength 20 and P.D. . Calculate the area of plates.
1
2
3
4
Explanation:
: Given that, Capacitance, Dielectric constant, Potential, volt Dielectric strength Let, separation between plates We know, Then, Capacitance
[AIIMS-25.05.2019(M) Shift-1]
Capacitance
166005
A capacitor of capacitance have dielectric slab of dielectric constant 2.5 , electric field strength and potential difference 30 volt. Calculate the area of plate.
1
2
3
4
Explanation:
: Given, Capacitance, Dielectric constant, Electric field strength, We know,
[AIIMS-25.05.2019(E)]**# Shift-2]
Capacitance
166006
When capacitor is fully charged, find current drawn from the cell.
1
2
3
4
Explanation:
: When capacitor is fully charged then it becomes open circuit hence no current flow. so the equivalent circuit is- Hence, current draw from the cell-
166002
A parallel plate capacitor has area separated by 3 dielectric slabs. Their relative permittivity is and thickness is , respectively. The capacitance is
1 Farad
2 Farad
3 Farad
4 Farad
Explanation:
: Given, Area (A) And Then, equivalent capacitance,
[COMEDK 2015]
Capacitance
166003
A capacitor is connected to a battery of voltage V. Now a dielectric slab of dielectric constant is completely inserted between the plates, then the final charge on the capacitor will be:
1
2
3
4 zero
Explanation:
: Given that, Voltage Dielectric constant We know that, Then, the final charge on the capacitor Putting the value of from equation (i)
[AIIMS-26.05.2019(E)]**# Shift-2]
Capacitance
166004
A capacitor of capacitance having dielectric slab of , dielectric strength 20 and P.D. . Calculate the area of plates.
1
2
3
4
Explanation:
: Given that, Capacitance, Dielectric constant, Potential, volt Dielectric strength Let, separation between plates We know, Then, Capacitance
[AIIMS-25.05.2019(M) Shift-1]
Capacitance
166005
A capacitor of capacitance have dielectric slab of dielectric constant 2.5 , electric field strength and potential difference 30 volt. Calculate the area of plate.
1
2
3
4
Explanation:
: Given, Capacitance, Dielectric constant, Electric field strength, We know,
[AIIMS-25.05.2019(E)]**# Shift-2]
Capacitance
166006
When capacitor is fully charged, find current drawn from the cell.
1
2
3
4
Explanation:
: When capacitor is fully charged then it becomes open circuit hence no current flow. so the equivalent circuit is- Hence, current draw from the cell-
166002
A parallel plate capacitor has area separated by 3 dielectric slabs. Their relative permittivity is and thickness is , respectively. The capacitance is
1 Farad
2 Farad
3 Farad
4 Farad
Explanation:
: Given, Area (A) And Then, equivalent capacitance,
[COMEDK 2015]
Capacitance
166003
A capacitor is connected to a battery of voltage V. Now a dielectric slab of dielectric constant is completely inserted between the plates, then the final charge on the capacitor will be:
1
2
3
4 zero
Explanation:
: Given that, Voltage Dielectric constant We know that, Then, the final charge on the capacitor Putting the value of from equation (i)
[AIIMS-26.05.2019(E)]**# Shift-2]
Capacitance
166004
A capacitor of capacitance having dielectric slab of , dielectric strength 20 and P.D. . Calculate the area of plates.
1
2
3
4
Explanation:
: Given that, Capacitance, Dielectric constant, Potential, volt Dielectric strength Let, separation between plates We know, Then, Capacitance
[AIIMS-25.05.2019(M) Shift-1]
Capacitance
166005
A capacitor of capacitance have dielectric slab of dielectric constant 2.5 , electric field strength and potential difference 30 volt. Calculate the area of plate.
1
2
3
4
Explanation:
: Given, Capacitance, Dielectric constant, Electric field strength, We know,
[AIIMS-25.05.2019(E)]**# Shift-2]
Capacitance
166006
When capacitor is fully charged, find current drawn from the cell.
1
2
3
4
Explanation:
: When capacitor is fully charged then it becomes open circuit hence no current flow. so the equivalent circuit is- Hence, current draw from the cell-
166002
A parallel plate capacitor has area separated by 3 dielectric slabs. Their relative permittivity is and thickness is , respectively. The capacitance is
1 Farad
2 Farad
3 Farad
4 Farad
Explanation:
: Given, Area (A) And Then, equivalent capacitance,
[COMEDK 2015]
Capacitance
166003
A capacitor is connected to a battery of voltage V. Now a dielectric slab of dielectric constant is completely inserted between the plates, then the final charge on the capacitor will be:
1
2
3
4 zero
Explanation:
: Given that, Voltage Dielectric constant We know that, Then, the final charge on the capacitor Putting the value of from equation (i)
[AIIMS-26.05.2019(E)]**# Shift-2]
Capacitance
166004
A capacitor of capacitance having dielectric slab of , dielectric strength 20 and P.D. . Calculate the area of plates.
1
2
3
4
Explanation:
: Given that, Capacitance, Dielectric constant, Potential, volt Dielectric strength Let, separation between plates We know, Then, Capacitance
[AIIMS-25.05.2019(M) Shift-1]
Capacitance
166005
A capacitor of capacitance have dielectric slab of dielectric constant 2.5 , electric field strength and potential difference 30 volt. Calculate the area of plate.
1
2
3
4
Explanation:
: Given, Capacitance, Dielectric constant, Electric field strength, We know,
[AIIMS-25.05.2019(E)]**# Shift-2]
Capacitance
166006
When capacitor is fully charged, find current drawn from the cell.
1
2
3
4
Explanation:
: When capacitor is fully charged then it becomes open circuit hence no current flow. so the equivalent circuit is- Hence, current draw from the cell-
166002
A parallel plate capacitor has area separated by 3 dielectric slabs. Their relative permittivity is and thickness is , respectively. The capacitance is
1 Farad
2 Farad
3 Farad
4 Farad
Explanation:
: Given, Area (A) And Then, equivalent capacitance,
[COMEDK 2015]
Capacitance
166003
A capacitor is connected to a battery of voltage V. Now a dielectric slab of dielectric constant is completely inserted between the plates, then the final charge on the capacitor will be:
1
2
3
4 zero
Explanation:
: Given that, Voltage Dielectric constant We know that, Then, the final charge on the capacitor Putting the value of from equation (i)
[AIIMS-26.05.2019(E)]**# Shift-2]
Capacitance
166004
A capacitor of capacitance having dielectric slab of , dielectric strength 20 and P.D. . Calculate the area of plates.
1
2
3
4
Explanation:
: Given that, Capacitance, Dielectric constant, Potential, volt Dielectric strength Let, separation between plates We know, Then, Capacitance
[AIIMS-25.05.2019(M) Shift-1]
Capacitance
166005
A capacitor of capacitance have dielectric slab of dielectric constant 2.5 , electric field strength and potential difference 30 volt. Calculate the area of plate.
1
2
3
4
Explanation:
: Given, Capacitance, Dielectric constant, Electric field strength, We know,
[AIIMS-25.05.2019(E)]**# Shift-2]
Capacitance
166006
When capacitor is fully charged, find current drawn from the cell.
1
2
3
4
Explanation:
: When capacitor is fully charged then it becomes open circuit hence no current flow. so the equivalent circuit is- Hence, current draw from the cell-