NEET Test Series from KOTA - 10 Papers In MS WORD
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Capacitance
165610
Instantaneous displacement current of $1.0 \mathrm{~A}$ in the space between the parallel plates of $1 \mu \mathrm{F}$ capacitor can be established by changing potential difference of
165611
A capacitor of $10 \mu \mathrm{F}$ is connected to a $10 \mathrm{~V}$ cell. The maximum charge on the capacitor will be
1 $1 \mu \mathrm{C}$
2 $10 \mu \mathrm{C}$
3 $100 \mu \mathrm{C}$
4 $1000 \mu \mathrm{C}$
Explanation:
: Given that, Capacitance, $\mathrm{C}=10 \mu \mathrm{F}$ Potential, $\mathrm{V}=10$ Volt We know, $\mathrm{Q}=\mathrm{CV}$ $\mathrm{Q}=10 \times 10$ $\mathrm{Q}=100 \mu \mathrm{C}$
[COMEDK 2012]
Capacitance
165612
Two capacitors $C_{1}$ and $C_{2}$ are charged to $120 \mathrm{~V}$ and $200 \mathrm{~V}$ respectively. When they are connected in parellel, it is found that potenial on each one of them is zero. Therefore,
1 $5 \mathrm{C}_{1}=3 \mathrm{C}_{2}$
2 $3 \mathrm{C}_{1}=5 \mathrm{C}_{2}$
3 $3 \mathrm{C}_{1}+5 \mathrm{C}_{2}=0$
4 $9 \mathrm{C}_{1}=4 \mathrm{C}_{2}$
Explanation:
: Given that, $\mathrm{Q}=\mathrm{CV}$ Two capacitors $\mathrm{C}_{1} \& \mathrm{C}_{2}$ $\mathrm{Q}_{1}=120 \mathrm{~V}, \quad \mathrm{Q}_{2}=200 \mathrm{~V}$ Potential can be made zero only when they are connected in parallel, $120 \mathrm{C}_{1}=200 \mathrm{C}_{2}$ $6 \mathrm{C}_{1}=10 \mathrm{C}_{2}$ $3 \mathrm{C}_{1}=5 \mathrm{C}_{2}$
[COMEDK 2020]
Capacitance
165613
A capacitor of capacitance $5 \mu \mathrm{F}$ is connected as shown in the figure. The internal resistance of the cell is $0.5 \Omega$. The amount of charge on the capacitor plate is-
1 zero
2 $5 \mu \mathrm{C}$
3 $10 \mu \mathrm{C}$
4 $25 \mu \mathrm{C}$
Explanation:
: Internal resistance $=0.5 \Omega$ Capacitor of capacitance $=5 \mu \mathrm{F}$ Net resistance of the circuit, $\mathrm{R}=1+1+0.5=2.5 \Omega$ Current flow from the cell $(\mathrm{I})=\frac{\mathrm{V}}{\mathrm{R}}=\frac{2.5}{2.5}=1 \mathrm{~A}$ Potential difference between two parallel plate; $\mathrm{V}=\mathrm{E}-\mathrm{Ir}=2.5-1 \times 0.5=2 \mathrm{~V}$ So, charge on the capacitor plates, $\mathrm{Q}=\mathrm{CV}=5 \times 2=10 \mu \mathrm{C}$
165610
Instantaneous displacement current of $1.0 \mathrm{~A}$ in the space between the parallel plates of $1 \mu \mathrm{F}$ capacitor can be established by changing potential difference of
165611
A capacitor of $10 \mu \mathrm{F}$ is connected to a $10 \mathrm{~V}$ cell. The maximum charge on the capacitor will be
1 $1 \mu \mathrm{C}$
2 $10 \mu \mathrm{C}$
3 $100 \mu \mathrm{C}$
4 $1000 \mu \mathrm{C}$
Explanation:
: Given that, Capacitance, $\mathrm{C}=10 \mu \mathrm{F}$ Potential, $\mathrm{V}=10$ Volt We know, $\mathrm{Q}=\mathrm{CV}$ $\mathrm{Q}=10 \times 10$ $\mathrm{Q}=100 \mu \mathrm{C}$
[COMEDK 2012]
Capacitance
165612
Two capacitors $C_{1}$ and $C_{2}$ are charged to $120 \mathrm{~V}$ and $200 \mathrm{~V}$ respectively. When they are connected in parellel, it is found that potenial on each one of them is zero. Therefore,
1 $5 \mathrm{C}_{1}=3 \mathrm{C}_{2}$
2 $3 \mathrm{C}_{1}=5 \mathrm{C}_{2}$
3 $3 \mathrm{C}_{1}+5 \mathrm{C}_{2}=0$
4 $9 \mathrm{C}_{1}=4 \mathrm{C}_{2}$
Explanation:
: Given that, $\mathrm{Q}=\mathrm{CV}$ Two capacitors $\mathrm{C}_{1} \& \mathrm{C}_{2}$ $\mathrm{Q}_{1}=120 \mathrm{~V}, \quad \mathrm{Q}_{2}=200 \mathrm{~V}$ Potential can be made zero only when they are connected in parallel, $120 \mathrm{C}_{1}=200 \mathrm{C}_{2}$ $6 \mathrm{C}_{1}=10 \mathrm{C}_{2}$ $3 \mathrm{C}_{1}=5 \mathrm{C}_{2}$
[COMEDK 2020]
Capacitance
165613
A capacitor of capacitance $5 \mu \mathrm{F}$ is connected as shown in the figure. The internal resistance of the cell is $0.5 \Omega$. The amount of charge on the capacitor plate is-
1 zero
2 $5 \mu \mathrm{C}$
3 $10 \mu \mathrm{C}$
4 $25 \mu \mathrm{C}$
Explanation:
: Internal resistance $=0.5 \Omega$ Capacitor of capacitance $=5 \mu \mathrm{F}$ Net resistance of the circuit, $\mathrm{R}=1+1+0.5=2.5 \Omega$ Current flow from the cell $(\mathrm{I})=\frac{\mathrm{V}}{\mathrm{R}}=\frac{2.5}{2.5}=1 \mathrm{~A}$ Potential difference between two parallel plate; $\mathrm{V}=\mathrm{E}-\mathrm{Ir}=2.5-1 \times 0.5=2 \mathrm{~V}$ So, charge on the capacitor plates, $\mathrm{Q}=\mathrm{CV}=5 \times 2=10 \mu \mathrm{C}$
165610
Instantaneous displacement current of $1.0 \mathrm{~A}$ in the space between the parallel plates of $1 \mu \mathrm{F}$ capacitor can be established by changing potential difference of
165611
A capacitor of $10 \mu \mathrm{F}$ is connected to a $10 \mathrm{~V}$ cell. The maximum charge on the capacitor will be
1 $1 \mu \mathrm{C}$
2 $10 \mu \mathrm{C}$
3 $100 \mu \mathrm{C}$
4 $1000 \mu \mathrm{C}$
Explanation:
: Given that, Capacitance, $\mathrm{C}=10 \mu \mathrm{F}$ Potential, $\mathrm{V}=10$ Volt We know, $\mathrm{Q}=\mathrm{CV}$ $\mathrm{Q}=10 \times 10$ $\mathrm{Q}=100 \mu \mathrm{C}$
[COMEDK 2012]
Capacitance
165612
Two capacitors $C_{1}$ and $C_{2}$ are charged to $120 \mathrm{~V}$ and $200 \mathrm{~V}$ respectively. When they are connected in parellel, it is found that potenial on each one of them is zero. Therefore,
1 $5 \mathrm{C}_{1}=3 \mathrm{C}_{2}$
2 $3 \mathrm{C}_{1}=5 \mathrm{C}_{2}$
3 $3 \mathrm{C}_{1}+5 \mathrm{C}_{2}=0$
4 $9 \mathrm{C}_{1}=4 \mathrm{C}_{2}$
Explanation:
: Given that, $\mathrm{Q}=\mathrm{CV}$ Two capacitors $\mathrm{C}_{1} \& \mathrm{C}_{2}$ $\mathrm{Q}_{1}=120 \mathrm{~V}, \quad \mathrm{Q}_{2}=200 \mathrm{~V}$ Potential can be made zero only when they are connected in parallel, $120 \mathrm{C}_{1}=200 \mathrm{C}_{2}$ $6 \mathrm{C}_{1}=10 \mathrm{C}_{2}$ $3 \mathrm{C}_{1}=5 \mathrm{C}_{2}$
[COMEDK 2020]
Capacitance
165613
A capacitor of capacitance $5 \mu \mathrm{F}$ is connected as shown in the figure. The internal resistance of the cell is $0.5 \Omega$. The amount of charge on the capacitor plate is-
1 zero
2 $5 \mu \mathrm{C}$
3 $10 \mu \mathrm{C}$
4 $25 \mu \mathrm{C}$
Explanation:
: Internal resistance $=0.5 \Omega$ Capacitor of capacitance $=5 \mu \mathrm{F}$ Net resistance of the circuit, $\mathrm{R}=1+1+0.5=2.5 \Omega$ Current flow from the cell $(\mathrm{I})=\frac{\mathrm{V}}{\mathrm{R}}=\frac{2.5}{2.5}=1 \mathrm{~A}$ Potential difference between two parallel plate; $\mathrm{V}=\mathrm{E}-\mathrm{Ir}=2.5-1 \times 0.5=2 \mathrm{~V}$ So, charge on the capacitor plates, $\mathrm{Q}=\mathrm{CV}=5 \times 2=10 \mu \mathrm{C}$
165610
Instantaneous displacement current of $1.0 \mathrm{~A}$ in the space between the parallel plates of $1 \mu \mathrm{F}$ capacitor can be established by changing potential difference of
165611
A capacitor of $10 \mu \mathrm{F}$ is connected to a $10 \mathrm{~V}$ cell. The maximum charge on the capacitor will be
1 $1 \mu \mathrm{C}$
2 $10 \mu \mathrm{C}$
3 $100 \mu \mathrm{C}$
4 $1000 \mu \mathrm{C}$
Explanation:
: Given that, Capacitance, $\mathrm{C}=10 \mu \mathrm{F}$ Potential, $\mathrm{V}=10$ Volt We know, $\mathrm{Q}=\mathrm{CV}$ $\mathrm{Q}=10 \times 10$ $\mathrm{Q}=100 \mu \mathrm{C}$
[COMEDK 2012]
Capacitance
165612
Two capacitors $C_{1}$ and $C_{2}$ are charged to $120 \mathrm{~V}$ and $200 \mathrm{~V}$ respectively. When they are connected in parellel, it is found that potenial on each one of them is zero. Therefore,
1 $5 \mathrm{C}_{1}=3 \mathrm{C}_{2}$
2 $3 \mathrm{C}_{1}=5 \mathrm{C}_{2}$
3 $3 \mathrm{C}_{1}+5 \mathrm{C}_{2}=0$
4 $9 \mathrm{C}_{1}=4 \mathrm{C}_{2}$
Explanation:
: Given that, $\mathrm{Q}=\mathrm{CV}$ Two capacitors $\mathrm{C}_{1} \& \mathrm{C}_{2}$ $\mathrm{Q}_{1}=120 \mathrm{~V}, \quad \mathrm{Q}_{2}=200 \mathrm{~V}$ Potential can be made zero only when they are connected in parallel, $120 \mathrm{C}_{1}=200 \mathrm{C}_{2}$ $6 \mathrm{C}_{1}=10 \mathrm{C}_{2}$ $3 \mathrm{C}_{1}=5 \mathrm{C}_{2}$
[COMEDK 2020]
Capacitance
165613
A capacitor of capacitance $5 \mu \mathrm{F}$ is connected as shown in the figure. The internal resistance of the cell is $0.5 \Omega$. The amount of charge on the capacitor plate is-
1 zero
2 $5 \mu \mathrm{C}$
3 $10 \mu \mathrm{C}$
4 $25 \mu \mathrm{C}$
Explanation:
: Internal resistance $=0.5 \Omega$ Capacitor of capacitance $=5 \mu \mathrm{F}$ Net resistance of the circuit, $\mathrm{R}=1+1+0.5=2.5 \Omega$ Current flow from the cell $(\mathrm{I})=\frac{\mathrm{V}}{\mathrm{R}}=\frac{2.5}{2.5}=1 \mathrm{~A}$ Potential difference between two parallel plate; $\mathrm{V}=\mathrm{E}-\mathrm{Ir}=2.5-1 \times 0.5=2 \mathrm{~V}$ So, charge on the capacitor plates, $\mathrm{Q}=\mathrm{CV}=5 \times 2=10 \mu \mathrm{C}$
