165823
' $n$ ' identical capacitors are joined in parallel and are charged to potential ' $V$ '. Now they are separated and joined in series, then
1 the potential difference is ' $\mathrm{nV}$ ' and energy increase ' $n$ ' times.
2 the potential difference remains the same and energy increases ' $n$ ' times.
3 the potential difference and the total energy of the combination remain the same.
4 the potential difference becomes ' $\mathrm{nV}$ ' and energy remains the same.
Explanation:
: Energy in parallel connection, $\mathrm{U}_{\mathrm{P}}=\frac{1}{2}(\mathrm{nC}) \mathrm{V}_{\mathrm{P}}^{2}$ Energy in series connection, $\mathrm{U}_{\mathrm{S}}=\frac{1}{2}\left(\frac{\mathrm{C}}{\mathrm{n}}\right) \mathrm{V}_{\mathrm{S}}^{2}$ According to energy conservation, energy remains the same. $\mathrm{U}_{\mathrm{P}}=\mathrm{U}_{\mathrm{S}}$ $\frac{1}{2}(\mathrm{nC}) \mathrm{V}_{\mathrm{P}}^{2}=\frac{1}{2}\left(\frac{\mathrm{C}}{\mathrm{n}}\right) \mathrm{V}_{\mathrm{S}}^{2}$ $\mathrm{~V}_{\mathrm{S}}=\mathrm{nV}_{\mathrm{P}}$
[MHT-CET 2020]
Capacitance
165824
Three condensers of capacities $C_{1}, C_{2}, C_{3}$ are connected in series with a source of e.m.f. $V$. The potentials across the three condensers are in the ratio of
: Three condensers of capacities $\mathrm{C}_{1}, \mathrm{C}_{2}, \mathrm{C}_{3}$ are connected in series with a source of e.m.f $\mathrm{V}$. $\mathrm{Q}=\mathrm{C}_{1} \mathrm{~V}_{1}=\mathrm{C}_{2} \mathrm{~V}_{2}=\mathrm{C}_{3} \mathrm{~V}_{3}$ $\mathrm{~V}_{1}=\frac{\mathrm{Q}}{\mathrm{C}_{1}}, \mathrm{~V}_{2}=\frac{\mathrm{Q}}{\mathrm{C}_{2}}, \mathrm{~V}_{3}=\frac{\mathrm{Q}}{\mathrm{C}_{3}}$ $\therefore \quad \mathrm{Q}=1$ Ratio of potential across the three condensers, $\mathrm{V}_{1}: \mathrm{V}_{2}: \mathrm{V}_{3}=\frac{1}{\mathrm{C}_{1}}: \frac{1}{\mathrm{C}_{2}}: \frac{1}{\mathrm{C}_{3}}$
[MHT-CET 2020]
Capacitance
165825
Four capacitors of equal capacity have an equivalent capacitance $C_{1}$ when connected in series and an equivalent capacitance $C_{2}$ when connected in parallel. The ratio $\frac{\mathrm{C}_{2}}{\mathrm{C}_{1}}$, is
1 4
2 8
3 16
4 12
Explanation:
: Four capacitors of equal capacity have an equivalent capacitance $\mathrm{C}_{1}$ when connected in series, $\frac{1}{\mathrm{C}_{1}}=\frac{1}{\mathrm{C}}+\frac{1}{\mathrm{C}}+\frac{1}{\mathrm{C}}+\frac{1}{\mathrm{C}}$ $\mathrm{C}_{1}=\frac{\mathrm{C}}{4}$ Again, connected in parallel, $\mathrm{C}_{2}=\mathrm{C}+\mathrm{C}+\mathrm{C}+\mathrm{C}$ $\mathrm{C}_{2}=4 \mathrm{C}$ The ratio of $\frac{\mathrm{C}_{2}}{\mathrm{C}_{1}}=\frac{4 \mathrm{C}}{\mathrm{C}} \times 4$ $=16$
[MHT-CET 2020]
Capacitance
165826
Five capacitors each of capacity ' $C$ ' are connected as shown in figure. If their resultant capacity is $2 \mu \mathrm{F}$, then the capacity of each condenser is
1 $2.5 \mu \mathrm{F}$
2 $10 \mu \mathrm{F}$
3 $5 \mu \mathrm{F}$
4 $2 \mu \mathrm{F}$
Explanation:
: Given, resultant capacitor $\left(\mathrm{C}_{\mathrm{R}}\right)=2 \mu \mathrm{F}$ Five capacitor of capacity ' $\mathrm{C}$ ' are connected in series. $\frac{1}{C_{R}}=\frac{1}{C}+\frac{1}{C}+\frac{1}{C}+\frac{1}{C}+\frac{1}{C}$ $C_{R}=\frac{C}{5}$ $C=5 \times C_{R}=5 \times 2 \times 10^{-6}=10 \mu F$ The capacity of each condenser is $10 \mu \mathrm{F}$.
[MHT-CET 2020]
Capacitance
165827
A condenser of capacity ' $C{ }_{1}$ ' is charged to potential ' $\mathrm{V}_{1}$ ' and then disconnected. Uncharged capacitor of capacity ' $\mathrm{C}_{2}$ ' is connected in parallel with ' $C_{1}$ '. The resultant potential ' $V_{2}$ ' is
: Charge $(\mathrm{Q})=\mathrm{CV}$ $\mathrm{Q}_{1}=\mathrm{C}_{1} \mathrm{~V}_{1}$ Capacitor $\mathrm{C}_{1}$ and $\mathrm{C}_{2}$ are connected in parallel, $\mathrm{C}_{\mathrm{R}}=\mathrm{C}_{1}+\mathrm{C}_{2}$ Resultant potential $\left(V_{2}\right)=\frac{Q_{1}}{C_{R}}$ Putting the value of $Q_{1}, C_{R}$ from equation (i) \& (ii), $\mathrm{V}_{2}=\frac{\mathrm{C}_{1} \mathrm{~V}_{1}}{\mathrm{C}_{1}+\mathrm{C}_{2}}$
165823
' $n$ ' identical capacitors are joined in parallel and are charged to potential ' $V$ '. Now they are separated and joined in series, then
1 the potential difference is ' $\mathrm{nV}$ ' and energy increase ' $n$ ' times.
2 the potential difference remains the same and energy increases ' $n$ ' times.
3 the potential difference and the total energy of the combination remain the same.
4 the potential difference becomes ' $\mathrm{nV}$ ' and energy remains the same.
Explanation:
: Energy in parallel connection, $\mathrm{U}_{\mathrm{P}}=\frac{1}{2}(\mathrm{nC}) \mathrm{V}_{\mathrm{P}}^{2}$ Energy in series connection, $\mathrm{U}_{\mathrm{S}}=\frac{1}{2}\left(\frac{\mathrm{C}}{\mathrm{n}}\right) \mathrm{V}_{\mathrm{S}}^{2}$ According to energy conservation, energy remains the same. $\mathrm{U}_{\mathrm{P}}=\mathrm{U}_{\mathrm{S}}$ $\frac{1}{2}(\mathrm{nC}) \mathrm{V}_{\mathrm{P}}^{2}=\frac{1}{2}\left(\frac{\mathrm{C}}{\mathrm{n}}\right) \mathrm{V}_{\mathrm{S}}^{2}$ $\mathrm{~V}_{\mathrm{S}}=\mathrm{nV}_{\mathrm{P}}$
[MHT-CET 2020]
Capacitance
165824
Three condensers of capacities $C_{1}, C_{2}, C_{3}$ are connected in series with a source of e.m.f. $V$. The potentials across the three condensers are in the ratio of
: Three condensers of capacities $\mathrm{C}_{1}, \mathrm{C}_{2}, \mathrm{C}_{3}$ are connected in series with a source of e.m.f $\mathrm{V}$. $\mathrm{Q}=\mathrm{C}_{1} \mathrm{~V}_{1}=\mathrm{C}_{2} \mathrm{~V}_{2}=\mathrm{C}_{3} \mathrm{~V}_{3}$ $\mathrm{~V}_{1}=\frac{\mathrm{Q}}{\mathrm{C}_{1}}, \mathrm{~V}_{2}=\frac{\mathrm{Q}}{\mathrm{C}_{2}}, \mathrm{~V}_{3}=\frac{\mathrm{Q}}{\mathrm{C}_{3}}$ $\therefore \quad \mathrm{Q}=1$ Ratio of potential across the three condensers, $\mathrm{V}_{1}: \mathrm{V}_{2}: \mathrm{V}_{3}=\frac{1}{\mathrm{C}_{1}}: \frac{1}{\mathrm{C}_{2}}: \frac{1}{\mathrm{C}_{3}}$
[MHT-CET 2020]
Capacitance
165825
Four capacitors of equal capacity have an equivalent capacitance $C_{1}$ when connected in series and an equivalent capacitance $C_{2}$ when connected in parallel. The ratio $\frac{\mathrm{C}_{2}}{\mathrm{C}_{1}}$, is
1 4
2 8
3 16
4 12
Explanation:
: Four capacitors of equal capacity have an equivalent capacitance $\mathrm{C}_{1}$ when connected in series, $\frac{1}{\mathrm{C}_{1}}=\frac{1}{\mathrm{C}}+\frac{1}{\mathrm{C}}+\frac{1}{\mathrm{C}}+\frac{1}{\mathrm{C}}$ $\mathrm{C}_{1}=\frac{\mathrm{C}}{4}$ Again, connected in parallel, $\mathrm{C}_{2}=\mathrm{C}+\mathrm{C}+\mathrm{C}+\mathrm{C}$ $\mathrm{C}_{2}=4 \mathrm{C}$ The ratio of $\frac{\mathrm{C}_{2}}{\mathrm{C}_{1}}=\frac{4 \mathrm{C}}{\mathrm{C}} \times 4$ $=16$
[MHT-CET 2020]
Capacitance
165826
Five capacitors each of capacity ' $C$ ' are connected as shown in figure. If their resultant capacity is $2 \mu \mathrm{F}$, then the capacity of each condenser is
1 $2.5 \mu \mathrm{F}$
2 $10 \mu \mathrm{F}$
3 $5 \mu \mathrm{F}$
4 $2 \mu \mathrm{F}$
Explanation:
: Given, resultant capacitor $\left(\mathrm{C}_{\mathrm{R}}\right)=2 \mu \mathrm{F}$ Five capacitor of capacity ' $\mathrm{C}$ ' are connected in series. $\frac{1}{C_{R}}=\frac{1}{C}+\frac{1}{C}+\frac{1}{C}+\frac{1}{C}+\frac{1}{C}$ $C_{R}=\frac{C}{5}$ $C=5 \times C_{R}=5 \times 2 \times 10^{-6}=10 \mu F$ The capacity of each condenser is $10 \mu \mathrm{F}$.
[MHT-CET 2020]
Capacitance
165827
A condenser of capacity ' $C{ }_{1}$ ' is charged to potential ' $\mathrm{V}_{1}$ ' and then disconnected. Uncharged capacitor of capacity ' $\mathrm{C}_{2}$ ' is connected in parallel with ' $C_{1}$ '. The resultant potential ' $V_{2}$ ' is
: Charge $(\mathrm{Q})=\mathrm{CV}$ $\mathrm{Q}_{1}=\mathrm{C}_{1} \mathrm{~V}_{1}$ Capacitor $\mathrm{C}_{1}$ and $\mathrm{C}_{2}$ are connected in parallel, $\mathrm{C}_{\mathrm{R}}=\mathrm{C}_{1}+\mathrm{C}_{2}$ Resultant potential $\left(V_{2}\right)=\frac{Q_{1}}{C_{R}}$ Putting the value of $Q_{1}, C_{R}$ from equation (i) \& (ii), $\mathrm{V}_{2}=\frac{\mathrm{C}_{1} \mathrm{~V}_{1}}{\mathrm{C}_{1}+\mathrm{C}_{2}}$
165823
' $n$ ' identical capacitors are joined in parallel and are charged to potential ' $V$ '. Now they are separated and joined in series, then
1 the potential difference is ' $\mathrm{nV}$ ' and energy increase ' $n$ ' times.
2 the potential difference remains the same and energy increases ' $n$ ' times.
3 the potential difference and the total energy of the combination remain the same.
4 the potential difference becomes ' $\mathrm{nV}$ ' and energy remains the same.
Explanation:
: Energy in parallel connection, $\mathrm{U}_{\mathrm{P}}=\frac{1}{2}(\mathrm{nC}) \mathrm{V}_{\mathrm{P}}^{2}$ Energy in series connection, $\mathrm{U}_{\mathrm{S}}=\frac{1}{2}\left(\frac{\mathrm{C}}{\mathrm{n}}\right) \mathrm{V}_{\mathrm{S}}^{2}$ According to energy conservation, energy remains the same. $\mathrm{U}_{\mathrm{P}}=\mathrm{U}_{\mathrm{S}}$ $\frac{1}{2}(\mathrm{nC}) \mathrm{V}_{\mathrm{P}}^{2}=\frac{1}{2}\left(\frac{\mathrm{C}}{\mathrm{n}}\right) \mathrm{V}_{\mathrm{S}}^{2}$ $\mathrm{~V}_{\mathrm{S}}=\mathrm{nV}_{\mathrm{P}}$
[MHT-CET 2020]
Capacitance
165824
Three condensers of capacities $C_{1}, C_{2}, C_{3}$ are connected in series with a source of e.m.f. $V$. The potentials across the three condensers are in the ratio of
: Three condensers of capacities $\mathrm{C}_{1}, \mathrm{C}_{2}, \mathrm{C}_{3}$ are connected in series with a source of e.m.f $\mathrm{V}$. $\mathrm{Q}=\mathrm{C}_{1} \mathrm{~V}_{1}=\mathrm{C}_{2} \mathrm{~V}_{2}=\mathrm{C}_{3} \mathrm{~V}_{3}$ $\mathrm{~V}_{1}=\frac{\mathrm{Q}}{\mathrm{C}_{1}}, \mathrm{~V}_{2}=\frac{\mathrm{Q}}{\mathrm{C}_{2}}, \mathrm{~V}_{3}=\frac{\mathrm{Q}}{\mathrm{C}_{3}}$ $\therefore \quad \mathrm{Q}=1$ Ratio of potential across the three condensers, $\mathrm{V}_{1}: \mathrm{V}_{2}: \mathrm{V}_{3}=\frac{1}{\mathrm{C}_{1}}: \frac{1}{\mathrm{C}_{2}}: \frac{1}{\mathrm{C}_{3}}$
[MHT-CET 2020]
Capacitance
165825
Four capacitors of equal capacity have an equivalent capacitance $C_{1}$ when connected in series and an equivalent capacitance $C_{2}$ when connected in parallel. The ratio $\frac{\mathrm{C}_{2}}{\mathrm{C}_{1}}$, is
1 4
2 8
3 16
4 12
Explanation:
: Four capacitors of equal capacity have an equivalent capacitance $\mathrm{C}_{1}$ when connected in series, $\frac{1}{\mathrm{C}_{1}}=\frac{1}{\mathrm{C}}+\frac{1}{\mathrm{C}}+\frac{1}{\mathrm{C}}+\frac{1}{\mathrm{C}}$ $\mathrm{C}_{1}=\frac{\mathrm{C}}{4}$ Again, connected in parallel, $\mathrm{C}_{2}=\mathrm{C}+\mathrm{C}+\mathrm{C}+\mathrm{C}$ $\mathrm{C}_{2}=4 \mathrm{C}$ The ratio of $\frac{\mathrm{C}_{2}}{\mathrm{C}_{1}}=\frac{4 \mathrm{C}}{\mathrm{C}} \times 4$ $=16$
[MHT-CET 2020]
Capacitance
165826
Five capacitors each of capacity ' $C$ ' are connected as shown in figure. If their resultant capacity is $2 \mu \mathrm{F}$, then the capacity of each condenser is
1 $2.5 \mu \mathrm{F}$
2 $10 \mu \mathrm{F}$
3 $5 \mu \mathrm{F}$
4 $2 \mu \mathrm{F}$
Explanation:
: Given, resultant capacitor $\left(\mathrm{C}_{\mathrm{R}}\right)=2 \mu \mathrm{F}$ Five capacitor of capacity ' $\mathrm{C}$ ' are connected in series. $\frac{1}{C_{R}}=\frac{1}{C}+\frac{1}{C}+\frac{1}{C}+\frac{1}{C}+\frac{1}{C}$ $C_{R}=\frac{C}{5}$ $C=5 \times C_{R}=5 \times 2 \times 10^{-6}=10 \mu F$ The capacity of each condenser is $10 \mu \mathrm{F}$.
[MHT-CET 2020]
Capacitance
165827
A condenser of capacity ' $C{ }_{1}$ ' is charged to potential ' $\mathrm{V}_{1}$ ' and then disconnected. Uncharged capacitor of capacity ' $\mathrm{C}_{2}$ ' is connected in parallel with ' $C_{1}$ '. The resultant potential ' $V_{2}$ ' is
: Charge $(\mathrm{Q})=\mathrm{CV}$ $\mathrm{Q}_{1}=\mathrm{C}_{1} \mathrm{~V}_{1}$ Capacitor $\mathrm{C}_{1}$ and $\mathrm{C}_{2}$ are connected in parallel, $\mathrm{C}_{\mathrm{R}}=\mathrm{C}_{1}+\mathrm{C}_{2}$ Resultant potential $\left(V_{2}\right)=\frac{Q_{1}}{C_{R}}$ Putting the value of $Q_{1}, C_{R}$ from equation (i) \& (ii), $\mathrm{V}_{2}=\frac{\mathrm{C}_{1} \mathrm{~V}_{1}}{\mathrm{C}_{1}+\mathrm{C}_{2}}$
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Capacitance
165823
' $n$ ' identical capacitors are joined in parallel and are charged to potential ' $V$ '. Now they are separated and joined in series, then
1 the potential difference is ' $\mathrm{nV}$ ' and energy increase ' $n$ ' times.
2 the potential difference remains the same and energy increases ' $n$ ' times.
3 the potential difference and the total energy of the combination remain the same.
4 the potential difference becomes ' $\mathrm{nV}$ ' and energy remains the same.
Explanation:
: Energy in parallel connection, $\mathrm{U}_{\mathrm{P}}=\frac{1}{2}(\mathrm{nC}) \mathrm{V}_{\mathrm{P}}^{2}$ Energy in series connection, $\mathrm{U}_{\mathrm{S}}=\frac{1}{2}\left(\frac{\mathrm{C}}{\mathrm{n}}\right) \mathrm{V}_{\mathrm{S}}^{2}$ According to energy conservation, energy remains the same. $\mathrm{U}_{\mathrm{P}}=\mathrm{U}_{\mathrm{S}}$ $\frac{1}{2}(\mathrm{nC}) \mathrm{V}_{\mathrm{P}}^{2}=\frac{1}{2}\left(\frac{\mathrm{C}}{\mathrm{n}}\right) \mathrm{V}_{\mathrm{S}}^{2}$ $\mathrm{~V}_{\mathrm{S}}=\mathrm{nV}_{\mathrm{P}}$
[MHT-CET 2020]
Capacitance
165824
Three condensers of capacities $C_{1}, C_{2}, C_{3}$ are connected in series with a source of e.m.f. $V$. The potentials across the three condensers are in the ratio of
: Three condensers of capacities $\mathrm{C}_{1}, \mathrm{C}_{2}, \mathrm{C}_{3}$ are connected in series with a source of e.m.f $\mathrm{V}$. $\mathrm{Q}=\mathrm{C}_{1} \mathrm{~V}_{1}=\mathrm{C}_{2} \mathrm{~V}_{2}=\mathrm{C}_{3} \mathrm{~V}_{3}$ $\mathrm{~V}_{1}=\frac{\mathrm{Q}}{\mathrm{C}_{1}}, \mathrm{~V}_{2}=\frac{\mathrm{Q}}{\mathrm{C}_{2}}, \mathrm{~V}_{3}=\frac{\mathrm{Q}}{\mathrm{C}_{3}}$ $\therefore \quad \mathrm{Q}=1$ Ratio of potential across the three condensers, $\mathrm{V}_{1}: \mathrm{V}_{2}: \mathrm{V}_{3}=\frac{1}{\mathrm{C}_{1}}: \frac{1}{\mathrm{C}_{2}}: \frac{1}{\mathrm{C}_{3}}$
[MHT-CET 2020]
Capacitance
165825
Four capacitors of equal capacity have an equivalent capacitance $C_{1}$ when connected in series and an equivalent capacitance $C_{2}$ when connected in parallel. The ratio $\frac{\mathrm{C}_{2}}{\mathrm{C}_{1}}$, is
1 4
2 8
3 16
4 12
Explanation:
: Four capacitors of equal capacity have an equivalent capacitance $\mathrm{C}_{1}$ when connected in series, $\frac{1}{\mathrm{C}_{1}}=\frac{1}{\mathrm{C}}+\frac{1}{\mathrm{C}}+\frac{1}{\mathrm{C}}+\frac{1}{\mathrm{C}}$ $\mathrm{C}_{1}=\frac{\mathrm{C}}{4}$ Again, connected in parallel, $\mathrm{C}_{2}=\mathrm{C}+\mathrm{C}+\mathrm{C}+\mathrm{C}$ $\mathrm{C}_{2}=4 \mathrm{C}$ The ratio of $\frac{\mathrm{C}_{2}}{\mathrm{C}_{1}}=\frac{4 \mathrm{C}}{\mathrm{C}} \times 4$ $=16$
[MHT-CET 2020]
Capacitance
165826
Five capacitors each of capacity ' $C$ ' are connected as shown in figure. If their resultant capacity is $2 \mu \mathrm{F}$, then the capacity of each condenser is
1 $2.5 \mu \mathrm{F}$
2 $10 \mu \mathrm{F}$
3 $5 \mu \mathrm{F}$
4 $2 \mu \mathrm{F}$
Explanation:
: Given, resultant capacitor $\left(\mathrm{C}_{\mathrm{R}}\right)=2 \mu \mathrm{F}$ Five capacitor of capacity ' $\mathrm{C}$ ' are connected in series. $\frac{1}{C_{R}}=\frac{1}{C}+\frac{1}{C}+\frac{1}{C}+\frac{1}{C}+\frac{1}{C}$ $C_{R}=\frac{C}{5}$ $C=5 \times C_{R}=5 \times 2 \times 10^{-6}=10 \mu F$ The capacity of each condenser is $10 \mu \mathrm{F}$.
[MHT-CET 2020]
Capacitance
165827
A condenser of capacity ' $C{ }_{1}$ ' is charged to potential ' $\mathrm{V}_{1}$ ' and then disconnected. Uncharged capacitor of capacity ' $\mathrm{C}_{2}$ ' is connected in parallel with ' $C_{1}$ '. The resultant potential ' $V_{2}$ ' is
: Charge $(\mathrm{Q})=\mathrm{CV}$ $\mathrm{Q}_{1}=\mathrm{C}_{1} \mathrm{~V}_{1}$ Capacitor $\mathrm{C}_{1}$ and $\mathrm{C}_{2}$ are connected in parallel, $\mathrm{C}_{\mathrm{R}}=\mathrm{C}_{1}+\mathrm{C}_{2}$ Resultant potential $\left(V_{2}\right)=\frac{Q_{1}}{C_{R}}$ Putting the value of $Q_{1}, C_{R}$ from equation (i) \& (ii), $\mathrm{V}_{2}=\frac{\mathrm{C}_{1} \mathrm{~V}_{1}}{\mathrm{C}_{1}+\mathrm{C}_{2}}$
165823
' $n$ ' identical capacitors are joined in parallel and are charged to potential ' $V$ '. Now they are separated and joined in series, then
1 the potential difference is ' $\mathrm{nV}$ ' and energy increase ' $n$ ' times.
2 the potential difference remains the same and energy increases ' $n$ ' times.
3 the potential difference and the total energy of the combination remain the same.
4 the potential difference becomes ' $\mathrm{nV}$ ' and energy remains the same.
Explanation:
: Energy in parallel connection, $\mathrm{U}_{\mathrm{P}}=\frac{1}{2}(\mathrm{nC}) \mathrm{V}_{\mathrm{P}}^{2}$ Energy in series connection, $\mathrm{U}_{\mathrm{S}}=\frac{1}{2}\left(\frac{\mathrm{C}}{\mathrm{n}}\right) \mathrm{V}_{\mathrm{S}}^{2}$ According to energy conservation, energy remains the same. $\mathrm{U}_{\mathrm{P}}=\mathrm{U}_{\mathrm{S}}$ $\frac{1}{2}(\mathrm{nC}) \mathrm{V}_{\mathrm{P}}^{2}=\frac{1}{2}\left(\frac{\mathrm{C}}{\mathrm{n}}\right) \mathrm{V}_{\mathrm{S}}^{2}$ $\mathrm{~V}_{\mathrm{S}}=\mathrm{nV}_{\mathrm{P}}$
[MHT-CET 2020]
Capacitance
165824
Three condensers of capacities $C_{1}, C_{2}, C_{3}$ are connected in series with a source of e.m.f. $V$. The potentials across the three condensers are in the ratio of
: Three condensers of capacities $\mathrm{C}_{1}, \mathrm{C}_{2}, \mathrm{C}_{3}$ are connected in series with a source of e.m.f $\mathrm{V}$. $\mathrm{Q}=\mathrm{C}_{1} \mathrm{~V}_{1}=\mathrm{C}_{2} \mathrm{~V}_{2}=\mathrm{C}_{3} \mathrm{~V}_{3}$ $\mathrm{~V}_{1}=\frac{\mathrm{Q}}{\mathrm{C}_{1}}, \mathrm{~V}_{2}=\frac{\mathrm{Q}}{\mathrm{C}_{2}}, \mathrm{~V}_{3}=\frac{\mathrm{Q}}{\mathrm{C}_{3}}$ $\therefore \quad \mathrm{Q}=1$ Ratio of potential across the three condensers, $\mathrm{V}_{1}: \mathrm{V}_{2}: \mathrm{V}_{3}=\frac{1}{\mathrm{C}_{1}}: \frac{1}{\mathrm{C}_{2}}: \frac{1}{\mathrm{C}_{3}}$
[MHT-CET 2020]
Capacitance
165825
Four capacitors of equal capacity have an equivalent capacitance $C_{1}$ when connected in series and an equivalent capacitance $C_{2}$ when connected in parallel. The ratio $\frac{\mathrm{C}_{2}}{\mathrm{C}_{1}}$, is
1 4
2 8
3 16
4 12
Explanation:
: Four capacitors of equal capacity have an equivalent capacitance $\mathrm{C}_{1}$ when connected in series, $\frac{1}{\mathrm{C}_{1}}=\frac{1}{\mathrm{C}}+\frac{1}{\mathrm{C}}+\frac{1}{\mathrm{C}}+\frac{1}{\mathrm{C}}$ $\mathrm{C}_{1}=\frac{\mathrm{C}}{4}$ Again, connected in parallel, $\mathrm{C}_{2}=\mathrm{C}+\mathrm{C}+\mathrm{C}+\mathrm{C}$ $\mathrm{C}_{2}=4 \mathrm{C}$ The ratio of $\frac{\mathrm{C}_{2}}{\mathrm{C}_{1}}=\frac{4 \mathrm{C}}{\mathrm{C}} \times 4$ $=16$
[MHT-CET 2020]
Capacitance
165826
Five capacitors each of capacity ' $C$ ' are connected as shown in figure. If their resultant capacity is $2 \mu \mathrm{F}$, then the capacity of each condenser is
1 $2.5 \mu \mathrm{F}$
2 $10 \mu \mathrm{F}$
3 $5 \mu \mathrm{F}$
4 $2 \mu \mathrm{F}$
Explanation:
: Given, resultant capacitor $\left(\mathrm{C}_{\mathrm{R}}\right)=2 \mu \mathrm{F}$ Five capacitor of capacity ' $\mathrm{C}$ ' are connected in series. $\frac{1}{C_{R}}=\frac{1}{C}+\frac{1}{C}+\frac{1}{C}+\frac{1}{C}+\frac{1}{C}$ $C_{R}=\frac{C}{5}$ $C=5 \times C_{R}=5 \times 2 \times 10^{-6}=10 \mu F$ The capacity of each condenser is $10 \mu \mathrm{F}$.
[MHT-CET 2020]
Capacitance
165827
A condenser of capacity ' $C{ }_{1}$ ' is charged to potential ' $\mathrm{V}_{1}$ ' and then disconnected. Uncharged capacitor of capacity ' $\mathrm{C}_{2}$ ' is connected in parallel with ' $C_{1}$ '. The resultant potential ' $V_{2}$ ' is
: Charge $(\mathrm{Q})=\mathrm{CV}$ $\mathrm{Q}_{1}=\mathrm{C}_{1} \mathrm{~V}_{1}$ Capacitor $\mathrm{C}_{1}$ and $\mathrm{C}_{2}$ are connected in parallel, $\mathrm{C}_{\mathrm{R}}=\mathrm{C}_{1}+\mathrm{C}_{2}$ Resultant potential $\left(V_{2}\right)=\frac{Q_{1}}{C_{R}}$ Putting the value of $Q_{1}, C_{R}$ from equation (i) \& (ii), $\mathrm{V}_{2}=\frac{\mathrm{C}_{1} \mathrm{~V}_{1}}{\mathrm{C}_{1}+\mathrm{C}_{2}}$
