165643
In a parallel plate capacitor, if $10^{12}$ electrons pass from one plate to another, a potential difference of $10 \mathrm{~V}$ is developed across the plates. The capacitance of the capacitor is
165644
In the given figure, the capacitors $C_{1}, C_{3}, C_{4}, C_{5}$ have a capacitance $4 \mu \mathrm{F}$ each. If the capacitor $\mathrm{C}_{2}$ has a capacitance $10 \mu \mathrm{F}$, then effective capacitance between $A$ and $B$ will be
1 $2 \mu \mathrm{F}$
2 $6 \mu \mathrm{F}$
3 $4 \mu \mathrm{F}$
4 $8 \mu \mathrm{F}$
Explanation:
: The given network can be replaced, $\frac{\mathrm{C}_{4}}{\mathrm{C}_{3}}=\frac{\mathrm{C}_{1}}{\mathrm{C}_{5}}$ No charge flow through $\mathrm{C}_{2}$. So, it is not considered - Hence, capacitors $\mathrm{C}_{1}$ and $\mathrm{C}_{5}$ are in series. $\frac{1}{C^{\prime}} =\frac{1}{C_{1}}+\frac{1}{C_{5}}$ $C^{\prime} =\frac{C_{1} C_{5}}{C_{1}+C_{5}}$ $C^{\prime} =\frac{4 \times 4}{4+4}$ $C^{\prime} =2 \mu F$ Similarly $\mathrm{C}_{4}$ and $\mathrm{C}_{3}$ are in series. Therefore, $\mathrm{C}^{\prime \prime}=\frac{\mathrm{C}_{3} \mathrm{C}_{4}}{\mathrm{C}_{3}+\mathrm{C}_{4}}=\frac{4 \times 4}{4+4}=2 \mu \mathrm{F}$ Also, $\mathrm{C}^{\prime}$ and $\mathrm{C}^{\prime \prime}$ are in parallel configuration. Hence, effective capacitance between $\mathrm{A}$ and $\mathrm{B}$ is, $\mathrm{C}_{\mathrm{AB}}=2+2=4 \mu \mathrm{F}$
[Karnataka CET - 2004
Capacitance
165645
Capacity of a parallel plate capacitor becomes $\frac{4}{3}$ times its original value when a dielectric slab of thickness $t=\frac{d}{3}$ is inserted between the plates of the capacitor, of separation ' $d$ '. Dielectric constant of the slab is
165646
A capacitor of capacitance $4 \mu \mathrm{F}$ is charged to a potential of $100 \mathrm{~V}$. It is then disconnected from the battery and connected in parallel with another capacitor $C_{2}$. If their common potential is 40 volts, then the value of $C_{2}$ is
165643
In a parallel plate capacitor, if $10^{12}$ electrons pass from one plate to another, a potential difference of $10 \mathrm{~V}$ is developed across the plates. The capacitance of the capacitor is
165644
In the given figure, the capacitors $C_{1}, C_{3}, C_{4}, C_{5}$ have a capacitance $4 \mu \mathrm{F}$ each. If the capacitor $\mathrm{C}_{2}$ has a capacitance $10 \mu \mathrm{F}$, then effective capacitance between $A$ and $B$ will be
1 $2 \mu \mathrm{F}$
2 $6 \mu \mathrm{F}$
3 $4 \mu \mathrm{F}$
4 $8 \mu \mathrm{F}$
Explanation:
: The given network can be replaced, $\frac{\mathrm{C}_{4}}{\mathrm{C}_{3}}=\frac{\mathrm{C}_{1}}{\mathrm{C}_{5}}$ No charge flow through $\mathrm{C}_{2}$. So, it is not considered - Hence, capacitors $\mathrm{C}_{1}$ and $\mathrm{C}_{5}$ are in series. $\frac{1}{C^{\prime}} =\frac{1}{C_{1}}+\frac{1}{C_{5}}$ $C^{\prime} =\frac{C_{1} C_{5}}{C_{1}+C_{5}}$ $C^{\prime} =\frac{4 \times 4}{4+4}$ $C^{\prime} =2 \mu F$ Similarly $\mathrm{C}_{4}$ and $\mathrm{C}_{3}$ are in series. Therefore, $\mathrm{C}^{\prime \prime}=\frac{\mathrm{C}_{3} \mathrm{C}_{4}}{\mathrm{C}_{3}+\mathrm{C}_{4}}=\frac{4 \times 4}{4+4}=2 \mu \mathrm{F}$ Also, $\mathrm{C}^{\prime}$ and $\mathrm{C}^{\prime \prime}$ are in parallel configuration. Hence, effective capacitance between $\mathrm{A}$ and $\mathrm{B}$ is, $\mathrm{C}_{\mathrm{AB}}=2+2=4 \mu \mathrm{F}$
[Karnataka CET - 2004
Capacitance
165645
Capacity of a parallel plate capacitor becomes $\frac{4}{3}$ times its original value when a dielectric slab of thickness $t=\frac{d}{3}$ is inserted between the plates of the capacitor, of separation ' $d$ '. Dielectric constant of the slab is
165646
A capacitor of capacitance $4 \mu \mathrm{F}$ is charged to a potential of $100 \mathrm{~V}$. It is then disconnected from the battery and connected in parallel with another capacitor $C_{2}$. If their common potential is 40 volts, then the value of $C_{2}$ is
165643
In a parallel plate capacitor, if $10^{12}$ electrons pass from one plate to another, a potential difference of $10 \mathrm{~V}$ is developed across the plates. The capacitance of the capacitor is
165644
In the given figure, the capacitors $C_{1}, C_{3}, C_{4}, C_{5}$ have a capacitance $4 \mu \mathrm{F}$ each. If the capacitor $\mathrm{C}_{2}$ has a capacitance $10 \mu \mathrm{F}$, then effective capacitance between $A$ and $B$ will be
1 $2 \mu \mathrm{F}$
2 $6 \mu \mathrm{F}$
3 $4 \mu \mathrm{F}$
4 $8 \mu \mathrm{F}$
Explanation:
: The given network can be replaced, $\frac{\mathrm{C}_{4}}{\mathrm{C}_{3}}=\frac{\mathrm{C}_{1}}{\mathrm{C}_{5}}$ No charge flow through $\mathrm{C}_{2}$. So, it is not considered - Hence, capacitors $\mathrm{C}_{1}$ and $\mathrm{C}_{5}$ are in series. $\frac{1}{C^{\prime}} =\frac{1}{C_{1}}+\frac{1}{C_{5}}$ $C^{\prime} =\frac{C_{1} C_{5}}{C_{1}+C_{5}}$ $C^{\prime} =\frac{4 \times 4}{4+4}$ $C^{\prime} =2 \mu F$ Similarly $\mathrm{C}_{4}$ and $\mathrm{C}_{3}$ are in series. Therefore, $\mathrm{C}^{\prime \prime}=\frac{\mathrm{C}_{3} \mathrm{C}_{4}}{\mathrm{C}_{3}+\mathrm{C}_{4}}=\frac{4 \times 4}{4+4}=2 \mu \mathrm{F}$ Also, $\mathrm{C}^{\prime}$ and $\mathrm{C}^{\prime \prime}$ are in parallel configuration. Hence, effective capacitance between $\mathrm{A}$ and $\mathrm{B}$ is, $\mathrm{C}_{\mathrm{AB}}=2+2=4 \mu \mathrm{F}$
[Karnataka CET - 2004
Capacitance
165645
Capacity of a parallel plate capacitor becomes $\frac{4}{3}$ times its original value when a dielectric slab of thickness $t=\frac{d}{3}$ is inserted between the plates of the capacitor, of separation ' $d$ '. Dielectric constant of the slab is
165646
A capacitor of capacitance $4 \mu \mathrm{F}$ is charged to a potential of $100 \mathrm{~V}$. It is then disconnected from the battery and connected in parallel with another capacitor $C_{2}$. If their common potential is 40 volts, then the value of $C_{2}$ is
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Capacitance
165643
In a parallel plate capacitor, if $10^{12}$ electrons pass from one plate to another, a potential difference of $10 \mathrm{~V}$ is developed across the plates. The capacitance of the capacitor is
165644
In the given figure, the capacitors $C_{1}, C_{3}, C_{4}, C_{5}$ have a capacitance $4 \mu \mathrm{F}$ each. If the capacitor $\mathrm{C}_{2}$ has a capacitance $10 \mu \mathrm{F}$, then effective capacitance between $A$ and $B$ will be
1 $2 \mu \mathrm{F}$
2 $6 \mu \mathrm{F}$
3 $4 \mu \mathrm{F}$
4 $8 \mu \mathrm{F}$
Explanation:
: The given network can be replaced, $\frac{\mathrm{C}_{4}}{\mathrm{C}_{3}}=\frac{\mathrm{C}_{1}}{\mathrm{C}_{5}}$ No charge flow through $\mathrm{C}_{2}$. So, it is not considered - Hence, capacitors $\mathrm{C}_{1}$ and $\mathrm{C}_{5}$ are in series. $\frac{1}{C^{\prime}} =\frac{1}{C_{1}}+\frac{1}{C_{5}}$ $C^{\prime} =\frac{C_{1} C_{5}}{C_{1}+C_{5}}$ $C^{\prime} =\frac{4 \times 4}{4+4}$ $C^{\prime} =2 \mu F$ Similarly $\mathrm{C}_{4}$ and $\mathrm{C}_{3}$ are in series. Therefore, $\mathrm{C}^{\prime \prime}=\frac{\mathrm{C}_{3} \mathrm{C}_{4}}{\mathrm{C}_{3}+\mathrm{C}_{4}}=\frac{4 \times 4}{4+4}=2 \mu \mathrm{F}$ Also, $\mathrm{C}^{\prime}$ and $\mathrm{C}^{\prime \prime}$ are in parallel configuration. Hence, effective capacitance between $\mathrm{A}$ and $\mathrm{B}$ is, $\mathrm{C}_{\mathrm{AB}}=2+2=4 \mu \mathrm{F}$
[Karnataka CET - 2004
Capacitance
165645
Capacity of a parallel plate capacitor becomes $\frac{4}{3}$ times its original value when a dielectric slab of thickness $t=\frac{d}{3}$ is inserted between the plates of the capacitor, of separation ' $d$ '. Dielectric constant of the slab is
165646
A capacitor of capacitance $4 \mu \mathrm{F}$ is charged to a potential of $100 \mathrm{~V}$. It is then disconnected from the battery and connected in parallel with another capacitor $C_{2}$. If their common potential is 40 volts, then the value of $C_{2}$ is
