165810
The capacitance between the point $A$ and $B$ in the following figure.
1 $\frac{3}{8} \mu \mathrm{F}$
2 $\frac{9}{4} \mu \mathrm{F}$
3 $\frac{4}{5} \mu \mathrm{F}$
4 $2 \mu \mathrm{F}$
Explanation:
: From given figure - $\mathrm{C}_{1}$ and $\mathrm{C}_{2}$ are connected in series so, equivalent capacitance is $\frac{1}{\mathrm{C}^{\prime}} =\frac{1}{\mathrm{C}_{1}}+\frac{1}{\mathrm{C}_{2}}$ $\frac{1}{\mathrm{C}^{\prime}} =\frac{1}{1}+\frac{1}{3}$ $\mathrm{C}^{\prime} =\frac{3 \times 1}{3+1}$ $\mathrm{C}^{\prime} =\frac{3}{4} \mu \mathrm{F}$ $\mathrm{C}_{2}$ and $\mathrm{C}_{3}$ are connected in series, so equivalent capacitance. $\frac{1}{\mathrm{C}^{\prime \prime}} =\frac{1}{\mathrm{C}_{3}}+\frac{1}{\mathrm{C}_{4}}$ $\frac{1}{\mathrm{C}^{\prime \prime}} =\frac{1}{2}+\frac{1}{6}$ $\mathrm{C}^{\prime \prime} =\frac{6 \times 2}{8}=\frac{6}{4} \mu \mathrm{F}$ Again redrawn figure - $\mathrm{C}^{\prime}$ and $\mathrm{C}^{\prime \prime}$ are connected in parallel. So, the equivalent capacitance between point $\mathrm{A}$ to $\mathrm{B}$ is $\mathrm{C}_{\mathrm{eq}} =\mathrm{C}^{\prime}+\mathrm{C}^{\prime \prime}$ $=\frac{3}{4}+\frac{6}{4}$ $=\frac{9}{4} \mu \mathrm{F}$
[AP EAMCET-05.07.2022
Capacitance
165811
The four capacitors of $25 \mu F$ each are connected as shown in the figure below. If the D.C. voltmeter read $200 \mathrm{~V}$. The charge on each plate of the capacitor is ......... .
1 $2 \times 10^{-3} \mathrm{C}$
2 $5 \times 10^{-3} \mathrm{C}$
3 $2 \times 10^{-2} \mathrm{C}$
4 $5 \times 10^{-2} \mathrm{C}$
Explanation:
: The two capacitors on the left side are in parallel and potential across each equal to similarly the capacitors on right side are in parallel and potential across each equal to $\mathrm{V}$. Therefore charge on each plate of capacitors is $\mathrm{Q}=\mathrm{CV}$ $\mathrm{Q}=25 \times 10^{-6} \times 200$ $\mathrm{Q}=5 \times 10^{-3} \mathrm{C}$
[AP EAMCET-23.09.2020
Capacitance
165812
In the circuit diagram shown in the adjoining figure the resultant capacitance between $P$ and $Q$ is
1 $47 \mu \mathrm{F}$
2 $3 \mu \mathrm{F}$
3 $60 \mu \mathrm{F}$
4 $10 \mu \mathrm{F}$
Explanation:
: From given figure, Redrawn the circuit- Where, $\mathrm{C}=(3+2) \mu \mathrm{F}=5 \mu \mathrm{F}$ equivalent capacitance between $\mathrm{P}$ and $\mathrm{Q}$ are $\frac{1}{\mathrm{C}_{\mathrm{PQ}}}=\frac{1}{5}+\frac{1}{20}+\frac{1}{12}$ $\frac{1}{\mathrm{C}_{\mathrm{PQ}}}=\frac{12+3+5}{60}$ $\mathrm{C}_{\mathrm{PQ}}=\frac{60}{20}$ $\mathrm{C}_{\mathrm{PQ}}=3 \mu \mathrm{F}$
[AP EAMCET-24.09.2020
Capacitance
165813
The energy of a parallel plate capacitor when connected to a battery is $E$. With the battery still in connection, if the plates of the capacitor are separated, so that the distance between them is twice the original distance, then the electrostatic energy becomes
1 $2 \mathrm{E}$
2 $\frac{E}{4}$
3 $\frac{E}{2}$
4 $4 \mathrm{E}$
Explanation:
: We know, Electrostatic energy of capacitor $(E)=\frac{1}{2} \mathrm{CV}^{2}$ And, $\quad \mathrm{C} \propto \frac{1}{\mathrm{~d}}$ $\frac{\mathrm{C}_{1}}{\mathrm{C}_{2}}=\frac{\mathrm{d}_{2}}{\mathrm{~d}_{1}}=\frac{2 \mathrm{~d}}{\mathrm{~d}}=2$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{\mathrm{C}_{1}}{\mathrm{C}_{2}}=\frac{2}{1}$ $\mathrm{E}_{2}=\frac{\mathrm{E}_{1}}{2}$ Since, $\mathrm{E}_{1}=\mathrm{E}$ So, $\mathrm{E}_{2}=\frac{\mathrm{E}}{2}$
165810
The capacitance between the point $A$ and $B$ in the following figure.
1 $\frac{3}{8} \mu \mathrm{F}$
2 $\frac{9}{4} \mu \mathrm{F}$
3 $\frac{4}{5} \mu \mathrm{F}$
4 $2 \mu \mathrm{F}$
Explanation:
: From given figure - $\mathrm{C}_{1}$ and $\mathrm{C}_{2}$ are connected in series so, equivalent capacitance is $\frac{1}{\mathrm{C}^{\prime}} =\frac{1}{\mathrm{C}_{1}}+\frac{1}{\mathrm{C}_{2}}$ $\frac{1}{\mathrm{C}^{\prime}} =\frac{1}{1}+\frac{1}{3}$ $\mathrm{C}^{\prime} =\frac{3 \times 1}{3+1}$ $\mathrm{C}^{\prime} =\frac{3}{4} \mu \mathrm{F}$ $\mathrm{C}_{2}$ and $\mathrm{C}_{3}$ are connected in series, so equivalent capacitance. $\frac{1}{\mathrm{C}^{\prime \prime}} =\frac{1}{\mathrm{C}_{3}}+\frac{1}{\mathrm{C}_{4}}$ $\frac{1}{\mathrm{C}^{\prime \prime}} =\frac{1}{2}+\frac{1}{6}$ $\mathrm{C}^{\prime \prime} =\frac{6 \times 2}{8}=\frac{6}{4} \mu \mathrm{F}$ Again redrawn figure - $\mathrm{C}^{\prime}$ and $\mathrm{C}^{\prime \prime}$ are connected in parallel. So, the equivalent capacitance between point $\mathrm{A}$ to $\mathrm{B}$ is $\mathrm{C}_{\mathrm{eq}} =\mathrm{C}^{\prime}+\mathrm{C}^{\prime \prime}$ $=\frac{3}{4}+\frac{6}{4}$ $=\frac{9}{4} \mu \mathrm{F}$
[AP EAMCET-05.07.2022
Capacitance
165811
The four capacitors of $25 \mu F$ each are connected as shown in the figure below. If the D.C. voltmeter read $200 \mathrm{~V}$. The charge on each plate of the capacitor is ......... .
1 $2 \times 10^{-3} \mathrm{C}$
2 $5 \times 10^{-3} \mathrm{C}$
3 $2 \times 10^{-2} \mathrm{C}$
4 $5 \times 10^{-2} \mathrm{C}$
Explanation:
: The two capacitors on the left side are in parallel and potential across each equal to similarly the capacitors on right side are in parallel and potential across each equal to $\mathrm{V}$. Therefore charge on each plate of capacitors is $\mathrm{Q}=\mathrm{CV}$ $\mathrm{Q}=25 \times 10^{-6} \times 200$ $\mathrm{Q}=5 \times 10^{-3} \mathrm{C}$
[AP EAMCET-23.09.2020
Capacitance
165812
In the circuit diagram shown in the adjoining figure the resultant capacitance between $P$ and $Q$ is
1 $47 \mu \mathrm{F}$
2 $3 \mu \mathrm{F}$
3 $60 \mu \mathrm{F}$
4 $10 \mu \mathrm{F}$
Explanation:
: From given figure, Redrawn the circuit- Where, $\mathrm{C}=(3+2) \mu \mathrm{F}=5 \mu \mathrm{F}$ equivalent capacitance between $\mathrm{P}$ and $\mathrm{Q}$ are $\frac{1}{\mathrm{C}_{\mathrm{PQ}}}=\frac{1}{5}+\frac{1}{20}+\frac{1}{12}$ $\frac{1}{\mathrm{C}_{\mathrm{PQ}}}=\frac{12+3+5}{60}$ $\mathrm{C}_{\mathrm{PQ}}=\frac{60}{20}$ $\mathrm{C}_{\mathrm{PQ}}=3 \mu \mathrm{F}$
[AP EAMCET-24.09.2020
Capacitance
165813
The energy of a parallel plate capacitor when connected to a battery is $E$. With the battery still in connection, if the plates of the capacitor are separated, so that the distance between them is twice the original distance, then the electrostatic energy becomes
1 $2 \mathrm{E}$
2 $\frac{E}{4}$
3 $\frac{E}{2}$
4 $4 \mathrm{E}$
Explanation:
: We know, Electrostatic energy of capacitor $(E)=\frac{1}{2} \mathrm{CV}^{2}$ And, $\quad \mathrm{C} \propto \frac{1}{\mathrm{~d}}$ $\frac{\mathrm{C}_{1}}{\mathrm{C}_{2}}=\frac{\mathrm{d}_{2}}{\mathrm{~d}_{1}}=\frac{2 \mathrm{~d}}{\mathrm{~d}}=2$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{\mathrm{C}_{1}}{\mathrm{C}_{2}}=\frac{2}{1}$ $\mathrm{E}_{2}=\frac{\mathrm{E}_{1}}{2}$ Since, $\mathrm{E}_{1}=\mathrm{E}$ So, $\mathrm{E}_{2}=\frac{\mathrm{E}}{2}$
165810
The capacitance between the point $A$ and $B$ in the following figure.
1 $\frac{3}{8} \mu \mathrm{F}$
2 $\frac{9}{4} \mu \mathrm{F}$
3 $\frac{4}{5} \mu \mathrm{F}$
4 $2 \mu \mathrm{F}$
Explanation:
: From given figure - $\mathrm{C}_{1}$ and $\mathrm{C}_{2}$ are connected in series so, equivalent capacitance is $\frac{1}{\mathrm{C}^{\prime}} =\frac{1}{\mathrm{C}_{1}}+\frac{1}{\mathrm{C}_{2}}$ $\frac{1}{\mathrm{C}^{\prime}} =\frac{1}{1}+\frac{1}{3}$ $\mathrm{C}^{\prime} =\frac{3 \times 1}{3+1}$ $\mathrm{C}^{\prime} =\frac{3}{4} \mu \mathrm{F}$ $\mathrm{C}_{2}$ and $\mathrm{C}_{3}$ are connected in series, so equivalent capacitance. $\frac{1}{\mathrm{C}^{\prime \prime}} =\frac{1}{\mathrm{C}_{3}}+\frac{1}{\mathrm{C}_{4}}$ $\frac{1}{\mathrm{C}^{\prime \prime}} =\frac{1}{2}+\frac{1}{6}$ $\mathrm{C}^{\prime \prime} =\frac{6 \times 2}{8}=\frac{6}{4} \mu \mathrm{F}$ Again redrawn figure - $\mathrm{C}^{\prime}$ and $\mathrm{C}^{\prime \prime}$ are connected in parallel. So, the equivalent capacitance between point $\mathrm{A}$ to $\mathrm{B}$ is $\mathrm{C}_{\mathrm{eq}} =\mathrm{C}^{\prime}+\mathrm{C}^{\prime \prime}$ $=\frac{3}{4}+\frac{6}{4}$ $=\frac{9}{4} \mu \mathrm{F}$
[AP EAMCET-05.07.2022
Capacitance
165811
The four capacitors of $25 \mu F$ each are connected as shown in the figure below. If the D.C. voltmeter read $200 \mathrm{~V}$. The charge on each plate of the capacitor is ......... .
1 $2 \times 10^{-3} \mathrm{C}$
2 $5 \times 10^{-3} \mathrm{C}$
3 $2 \times 10^{-2} \mathrm{C}$
4 $5 \times 10^{-2} \mathrm{C}$
Explanation:
: The two capacitors on the left side are in parallel and potential across each equal to similarly the capacitors on right side are in parallel and potential across each equal to $\mathrm{V}$. Therefore charge on each plate of capacitors is $\mathrm{Q}=\mathrm{CV}$ $\mathrm{Q}=25 \times 10^{-6} \times 200$ $\mathrm{Q}=5 \times 10^{-3} \mathrm{C}$
[AP EAMCET-23.09.2020
Capacitance
165812
In the circuit diagram shown in the adjoining figure the resultant capacitance between $P$ and $Q$ is
1 $47 \mu \mathrm{F}$
2 $3 \mu \mathrm{F}$
3 $60 \mu \mathrm{F}$
4 $10 \mu \mathrm{F}$
Explanation:
: From given figure, Redrawn the circuit- Where, $\mathrm{C}=(3+2) \mu \mathrm{F}=5 \mu \mathrm{F}$ equivalent capacitance between $\mathrm{P}$ and $\mathrm{Q}$ are $\frac{1}{\mathrm{C}_{\mathrm{PQ}}}=\frac{1}{5}+\frac{1}{20}+\frac{1}{12}$ $\frac{1}{\mathrm{C}_{\mathrm{PQ}}}=\frac{12+3+5}{60}$ $\mathrm{C}_{\mathrm{PQ}}=\frac{60}{20}$ $\mathrm{C}_{\mathrm{PQ}}=3 \mu \mathrm{F}$
[AP EAMCET-24.09.2020
Capacitance
165813
The energy of a parallel plate capacitor when connected to a battery is $E$. With the battery still in connection, if the plates of the capacitor are separated, so that the distance between them is twice the original distance, then the electrostatic energy becomes
1 $2 \mathrm{E}$
2 $\frac{E}{4}$
3 $\frac{E}{2}$
4 $4 \mathrm{E}$
Explanation:
: We know, Electrostatic energy of capacitor $(E)=\frac{1}{2} \mathrm{CV}^{2}$ And, $\quad \mathrm{C} \propto \frac{1}{\mathrm{~d}}$ $\frac{\mathrm{C}_{1}}{\mathrm{C}_{2}}=\frac{\mathrm{d}_{2}}{\mathrm{~d}_{1}}=\frac{2 \mathrm{~d}}{\mathrm{~d}}=2$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{\mathrm{C}_{1}}{\mathrm{C}_{2}}=\frac{2}{1}$ $\mathrm{E}_{2}=\frac{\mathrm{E}_{1}}{2}$ Since, $\mathrm{E}_{1}=\mathrm{E}$ So, $\mathrm{E}_{2}=\frac{\mathrm{E}}{2}$
165810
The capacitance between the point $A$ and $B$ in the following figure.
1 $\frac{3}{8} \mu \mathrm{F}$
2 $\frac{9}{4} \mu \mathrm{F}$
3 $\frac{4}{5} \mu \mathrm{F}$
4 $2 \mu \mathrm{F}$
Explanation:
: From given figure - $\mathrm{C}_{1}$ and $\mathrm{C}_{2}$ are connected in series so, equivalent capacitance is $\frac{1}{\mathrm{C}^{\prime}} =\frac{1}{\mathrm{C}_{1}}+\frac{1}{\mathrm{C}_{2}}$ $\frac{1}{\mathrm{C}^{\prime}} =\frac{1}{1}+\frac{1}{3}$ $\mathrm{C}^{\prime} =\frac{3 \times 1}{3+1}$ $\mathrm{C}^{\prime} =\frac{3}{4} \mu \mathrm{F}$ $\mathrm{C}_{2}$ and $\mathrm{C}_{3}$ are connected in series, so equivalent capacitance. $\frac{1}{\mathrm{C}^{\prime \prime}} =\frac{1}{\mathrm{C}_{3}}+\frac{1}{\mathrm{C}_{4}}$ $\frac{1}{\mathrm{C}^{\prime \prime}} =\frac{1}{2}+\frac{1}{6}$ $\mathrm{C}^{\prime \prime} =\frac{6 \times 2}{8}=\frac{6}{4} \mu \mathrm{F}$ Again redrawn figure - $\mathrm{C}^{\prime}$ and $\mathrm{C}^{\prime \prime}$ are connected in parallel. So, the equivalent capacitance between point $\mathrm{A}$ to $\mathrm{B}$ is $\mathrm{C}_{\mathrm{eq}} =\mathrm{C}^{\prime}+\mathrm{C}^{\prime \prime}$ $=\frac{3}{4}+\frac{6}{4}$ $=\frac{9}{4} \mu \mathrm{F}$
[AP EAMCET-05.07.2022
Capacitance
165811
The four capacitors of $25 \mu F$ each are connected as shown in the figure below. If the D.C. voltmeter read $200 \mathrm{~V}$. The charge on each plate of the capacitor is ......... .
1 $2 \times 10^{-3} \mathrm{C}$
2 $5 \times 10^{-3} \mathrm{C}$
3 $2 \times 10^{-2} \mathrm{C}$
4 $5 \times 10^{-2} \mathrm{C}$
Explanation:
: The two capacitors on the left side are in parallel and potential across each equal to similarly the capacitors on right side are in parallel and potential across each equal to $\mathrm{V}$. Therefore charge on each plate of capacitors is $\mathrm{Q}=\mathrm{CV}$ $\mathrm{Q}=25 \times 10^{-6} \times 200$ $\mathrm{Q}=5 \times 10^{-3} \mathrm{C}$
[AP EAMCET-23.09.2020
Capacitance
165812
In the circuit diagram shown in the adjoining figure the resultant capacitance between $P$ and $Q$ is
1 $47 \mu \mathrm{F}$
2 $3 \mu \mathrm{F}$
3 $60 \mu \mathrm{F}$
4 $10 \mu \mathrm{F}$
Explanation:
: From given figure, Redrawn the circuit- Where, $\mathrm{C}=(3+2) \mu \mathrm{F}=5 \mu \mathrm{F}$ equivalent capacitance between $\mathrm{P}$ and $\mathrm{Q}$ are $\frac{1}{\mathrm{C}_{\mathrm{PQ}}}=\frac{1}{5}+\frac{1}{20}+\frac{1}{12}$ $\frac{1}{\mathrm{C}_{\mathrm{PQ}}}=\frac{12+3+5}{60}$ $\mathrm{C}_{\mathrm{PQ}}=\frac{60}{20}$ $\mathrm{C}_{\mathrm{PQ}}=3 \mu \mathrm{F}$
[AP EAMCET-24.09.2020
Capacitance
165813
The energy of a parallel plate capacitor when connected to a battery is $E$. With the battery still in connection, if the plates of the capacitor are separated, so that the distance between them is twice the original distance, then the electrostatic energy becomes
1 $2 \mathrm{E}$
2 $\frac{E}{4}$
3 $\frac{E}{2}$
4 $4 \mathrm{E}$
Explanation:
: We know, Electrostatic energy of capacitor $(E)=\frac{1}{2} \mathrm{CV}^{2}$ And, $\quad \mathrm{C} \propto \frac{1}{\mathrm{~d}}$ $\frac{\mathrm{C}_{1}}{\mathrm{C}_{2}}=\frac{\mathrm{d}_{2}}{\mathrm{~d}_{1}}=\frac{2 \mathrm{~d}}{\mathrm{~d}}=2$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{\mathrm{C}_{1}}{\mathrm{C}_{2}}=\frac{2}{1}$ $\mathrm{E}_{2}=\frac{\mathrm{E}_{1}}{2}$ Since, $\mathrm{E}_{1}=\mathrm{E}$ So, $\mathrm{E}_{2}=\frac{\mathrm{E}}{2}$
