165819
The equivalent capacitance between the points $A$ and $B$ in the following circuit is :
1 $1 \mu \mathrm{F}$
2 $2 \mu \mathrm{F}$
3 $4 \mu \mathrm{F}$
4 $8 \mu \mathrm{F}$
Explanation:
: Capacitor of capacitance $1.5 \mu \mathrm{F}$ and $1.5 \mu \mathrm{F}$ are connected in parallel. Hence their equivalent resistance is $3 \mu \mathrm{F}$. Now, three capacitor each of value $3 \mu \mathrm{F}$ are in series. Hence, their equivalent capacitance is given by, $\frac{1}{\mathrm{C}}=\frac{1}{3}+\frac{1}{3}+\frac{1}{3}$ $\mathrm{C}=1 \mu \mathrm{F}$
[BCECE-2005
Capacitance
165820
When two conductors of charges and potentials $C_{1}, V_{1}$ and $C_{2}, V_{2}$ respectively are joined, the common potential will be:
: When two conductors of charges and potentials $\mathrm{C}_{1}, \mathrm{~V}_{1}$ and $\mathrm{C}_{2}, \mathrm{~V}_{2}$ respectively are joined. Let the common potential will be $\mathrm{V}$. By the law of conservation of charge, $\mathrm{Q}_{\mathrm{i}}=\mathrm{C}_{1} \mathrm{~V}_{1}+\mathrm{C}_{2} \mathrm{~V}_{2}$ $\mathrm{Q}_{\mathrm{f}}=\mathrm{C}_{1} \mathrm{~V}+\mathrm{C}_{2} \mathrm{~V}$ $\mathrm{Q}_{\mathrm{i}}=\mathrm{Q}_{\mathrm{f}}$ $\mathrm{C}_{1} \mathrm{~V}_{1}+\mathrm{C}_{2} \mathrm{~V}_{2}=\mathrm{C}_{1} \mathrm{~V}+\mathrm{C}_{2} \mathrm{~V}$ $\mathrm{~V}=\frac{\mathrm{C}_{1} \mathrm{~V}_{1}+\mathrm{C}_{2} \mathrm{~V}_{2}}{\mathrm{C}_{1}+\mathrm{C}_{2}}$
[BCECE-2003]
Capacitance
165821
In the circuit shown, the potential difference across the $4.5 \mu \mathrm{F}$ capacitor is
1 6 volt
2 $\frac{8}{3}$ volt
3 4 volt
4 8 volt
Explanation:
: Since, $4.6 \mu \mathrm{F}$ and $\mathrm{C}_{\mathrm{R}}=9 \mu \mathrm{F}$ are in series. So, the charge on capacitors is same. $\therefore \mathrm{C}_{1} \mathrm{~V}_{1}=\mathrm{C}_{2} \mathrm{~V}_{2}$ $\Rightarrow 9 \mathrm{~V}_{1}=4.5 \mathrm{~V}_{2}$ $\Rightarrow \mathrm{V}_{2}=2 \mathrm{~V}_{1}$ Given, $V_{1}+V_{2}=12$ Volt $\mathrm{V}_{1}+2 \mathrm{~V}_{1}=12$ $\Rightarrow 3 \mathrm{~V}_{1}=12$ $\Rightarrow \mathrm{V}_{1}=4$ volt $\mathrm{V}_{2}($ Voltage on $4 \mu \mathrm{F})=2 \times 4=8 \mathrm{~V}$
[MHT-CET 2020
Capacitance
165822
The potential differences that must be applied across the parallel and series combination of 3 identical capacitors are such that the energy stored in them becomes the same. The ratio of potential difference in parallel to series combination is
1 $\frac{1}{3}$
2 $\frac{1}{4}$
3 $\frac{1}{8}$
4 $\frac{1}{6}$
Explanation:
: Three capacitors are connected in parallel $\mathrm{C}_{\mathrm{P}}=\mathrm{C}+\mathrm{C}+\mathrm{C}$ $\mathrm{C}_{\mathrm{P}}=3 \mathrm{C}$ Three capacitors are connected in series, $\frac{1}{\mathrm{C}_{\mathrm{S}}}=\frac{1}{\mathrm{C}}+\frac{1}{\mathrm{C}}+\frac{1}{\mathrm{C}}$ $\mathrm{C}_{\mathrm{S}}=\frac{\mathrm{C}}{3}$ Energy stored is same in both conditions, $\frac{1}{2} C_{P} V_{P}^{2}=\frac{1}{2} C_{S} V_{S}^{2}$ $\left(\frac{V_{P}}{V_{S}}\right)^{2}=\frac{C_{S}}{C_{P}}$ $\frac{V_{P}}{V_{S}}=\sqrt{\frac{C_{S}}{C_{P}}}$ Ratio of potential difference in parallel to series combination, $\frac{\mathrm{V}_{\mathrm{P}}}{\mathrm{V}_{\mathrm{S}}}=\sqrt{\frac{\mathrm{C}}{3 \times 3 \mathrm{C}}}=\frac{1}{3}$
165819
The equivalent capacitance between the points $A$ and $B$ in the following circuit is :
1 $1 \mu \mathrm{F}$
2 $2 \mu \mathrm{F}$
3 $4 \mu \mathrm{F}$
4 $8 \mu \mathrm{F}$
Explanation:
: Capacitor of capacitance $1.5 \mu \mathrm{F}$ and $1.5 \mu \mathrm{F}$ are connected in parallel. Hence their equivalent resistance is $3 \mu \mathrm{F}$. Now, three capacitor each of value $3 \mu \mathrm{F}$ are in series. Hence, their equivalent capacitance is given by, $\frac{1}{\mathrm{C}}=\frac{1}{3}+\frac{1}{3}+\frac{1}{3}$ $\mathrm{C}=1 \mu \mathrm{F}$
[BCECE-2005
Capacitance
165820
When two conductors of charges and potentials $C_{1}, V_{1}$ and $C_{2}, V_{2}$ respectively are joined, the common potential will be:
: When two conductors of charges and potentials $\mathrm{C}_{1}, \mathrm{~V}_{1}$ and $\mathrm{C}_{2}, \mathrm{~V}_{2}$ respectively are joined. Let the common potential will be $\mathrm{V}$. By the law of conservation of charge, $\mathrm{Q}_{\mathrm{i}}=\mathrm{C}_{1} \mathrm{~V}_{1}+\mathrm{C}_{2} \mathrm{~V}_{2}$ $\mathrm{Q}_{\mathrm{f}}=\mathrm{C}_{1} \mathrm{~V}+\mathrm{C}_{2} \mathrm{~V}$ $\mathrm{Q}_{\mathrm{i}}=\mathrm{Q}_{\mathrm{f}}$ $\mathrm{C}_{1} \mathrm{~V}_{1}+\mathrm{C}_{2} \mathrm{~V}_{2}=\mathrm{C}_{1} \mathrm{~V}+\mathrm{C}_{2} \mathrm{~V}$ $\mathrm{~V}=\frac{\mathrm{C}_{1} \mathrm{~V}_{1}+\mathrm{C}_{2} \mathrm{~V}_{2}}{\mathrm{C}_{1}+\mathrm{C}_{2}}$
[BCECE-2003]
Capacitance
165821
In the circuit shown, the potential difference across the $4.5 \mu \mathrm{F}$ capacitor is
1 6 volt
2 $\frac{8}{3}$ volt
3 4 volt
4 8 volt
Explanation:
: Since, $4.6 \mu \mathrm{F}$ and $\mathrm{C}_{\mathrm{R}}=9 \mu \mathrm{F}$ are in series. So, the charge on capacitors is same. $\therefore \mathrm{C}_{1} \mathrm{~V}_{1}=\mathrm{C}_{2} \mathrm{~V}_{2}$ $\Rightarrow 9 \mathrm{~V}_{1}=4.5 \mathrm{~V}_{2}$ $\Rightarrow \mathrm{V}_{2}=2 \mathrm{~V}_{1}$ Given, $V_{1}+V_{2}=12$ Volt $\mathrm{V}_{1}+2 \mathrm{~V}_{1}=12$ $\Rightarrow 3 \mathrm{~V}_{1}=12$ $\Rightarrow \mathrm{V}_{1}=4$ volt $\mathrm{V}_{2}($ Voltage on $4 \mu \mathrm{F})=2 \times 4=8 \mathrm{~V}$
[MHT-CET 2020
Capacitance
165822
The potential differences that must be applied across the parallel and series combination of 3 identical capacitors are such that the energy stored in them becomes the same. The ratio of potential difference in parallel to series combination is
1 $\frac{1}{3}$
2 $\frac{1}{4}$
3 $\frac{1}{8}$
4 $\frac{1}{6}$
Explanation:
: Three capacitors are connected in parallel $\mathrm{C}_{\mathrm{P}}=\mathrm{C}+\mathrm{C}+\mathrm{C}$ $\mathrm{C}_{\mathrm{P}}=3 \mathrm{C}$ Three capacitors are connected in series, $\frac{1}{\mathrm{C}_{\mathrm{S}}}=\frac{1}{\mathrm{C}}+\frac{1}{\mathrm{C}}+\frac{1}{\mathrm{C}}$ $\mathrm{C}_{\mathrm{S}}=\frac{\mathrm{C}}{3}$ Energy stored is same in both conditions, $\frac{1}{2} C_{P} V_{P}^{2}=\frac{1}{2} C_{S} V_{S}^{2}$ $\left(\frac{V_{P}}{V_{S}}\right)^{2}=\frac{C_{S}}{C_{P}}$ $\frac{V_{P}}{V_{S}}=\sqrt{\frac{C_{S}}{C_{P}}}$ Ratio of potential difference in parallel to series combination, $\frac{\mathrm{V}_{\mathrm{P}}}{\mathrm{V}_{\mathrm{S}}}=\sqrt{\frac{\mathrm{C}}{3 \times 3 \mathrm{C}}}=\frac{1}{3}$
165819
The equivalent capacitance between the points $A$ and $B$ in the following circuit is :
1 $1 \mu \mathrm{F}$
2 $2 \mu \mathrm{F}$
3 $4 \mu \mathrm{F}$
4 $8 \mu \mathrm{F}$
Explanation:
: Capacitor of capacitance $1.5 \mu \mathrm{F}$ and $1.5 \mu \mathrm{F}$ are connected in parallel. Hence their equivalent resistance is $3 \mu \mathrm{F}$. Now, three capacitor each of value $3 \mu \mathrm{F}$ are in series. Hence, their equivalent capacitance is given by, $\frac{1}{\mathrm{C}}=\frac{1}{3}+\frac{1}{3}+\frac{1}{3}$ $\mathrm{C}=1 \mu \mathrm{F}$
[BCECE-2005
Capacitance
165820
When two conductors of charges and potentials $C_{1}, V_{1}$ and $C_{2}, V_{2}$ respectively are joined, the common potential will be:
: When two conductors of charges and potentials $\mathrm{C}_{1}, \mathrm{~V}_{1}$ and $\mathrm{C}_{2}, \mathrm{~V}_{2}$ respectively are joined. Let the common potential will be $\mathrm{V}$. By the law of conservation of charge, $\mathrm{Q}_{\mathrm{i}}=\mathrm{C}_{1} \mathrm{~V}_{1}+\mathrm{C}_{2} \mathrm{~V}_{2}$ $\mathrm{Q}_{\mathrm{f}}=\mathrm{C}_{1} \mathrm{~V}+\mathrm{C}_{2} \mathrm{~V}$ $\mathrm{Q}_{\mathrm{i}}=\mathrm{Q}_{\mathrm{f}}$ $\mathrm{C}_{1} \mathrm{~V}_{1}+\mathrm{C}_{2} \mathrm{~V}_{2}=\mathrm{C}_{1} \mathrm{~V}+\mathrm{C}_{2} \mathrm{~V}$ $\mathrm{~V}=\frac{\mathrm{C}_{1} \mathrm{~V}_{1}+\mathrm{C}_{2} \mathrm{~V}_{2}}{\mathrm{C}_{1}+\mathrm{C}_{2}}$
[BCECE-2003]
Capacitance
165821
In the circuit shown, the potential difference across the $4.5 \mu \mathrm{F}$ capacitor is
1 6 volt
2 $\frac{8}{3}$ volt
3 4 volt
4 8 volt
Explanation:
: Since, $4.6 \mu \mathrm{F}$ and $\mathrm{C}_{\mathrm{R}}=9 \mu \mathrm{F}$ are in series. So, the charge on capacitors is same. $\therefore \mathrm{C}_{1} \mathrm{~V}_{1}=\mathrm{C}_{2} \mathrm{~V}_{2}$ $\Rightarrow 9 \mathrm{~V}_{1}=4.5 \mathrm{~V}_{2}$ $\Rightarrow \mathrm{V}_{2}=2 \mathrm{~V}_{1}$ Given, $V_{1}+V_{2}=12$ Volt $\mathrm{V}_{1}+2 \mathrm{~V}_{1}=12$ $\Rightarrow 3 \mathrm{~V}_{1}=12$ $\Rightarrow \mathrm{V}_{1}=4$ volt $\mathrm{V}_{2}($ Voltage on $4 \mu \mathrm{F})=2 \times 4=8 \mathrm{~V}$
[MHT-CET 2020
Capacitance
165822
The potential differences that must be applied across the parallel and series combination of 3 identical capacitors are such that the energy stored in them becomes the same. The ratio of potential difference in parallel to series combination is
1 $\frac{1}{3}$
2 $\frac{1}{4}$
3 $\frac{1}{8}$
4 $\frac{1}{6}$
Explanation:
: Three capacitors are connected in parallel $\mathrm{C}_{\mathrm{P}}=\mathrm{C}+\mathrm{C}+\mathrm{C}$ $\mathrm{C}_{\mathrm{P}}=3 \mathrm{C}$ Three capacitors are connected in series, $\frac{1}{\mathrm{C}_{\mathrm{S}}}=\frac{1}{\mathrm{C}}+\frac{1}{\mathrm{C}}+\frac{1}{\mathrm{C}}$ $\mathrm{C}_{\mathrm{S}}=\frac{\mathrm{C}}{3}$ Energy stored is same in both conditions, $\frac{1}{2} C_{P} V_{P}^{2}=\frac{1}{2} C_{S} V_{S}^{2}$ $\left(\frac{V_{P}}{V_{S}}\right)^{2}=\frac{C_{S}}{C_{P}}$ $\frac{V_{P}}{V_{S}}=\sqrt{\frac{C_{S}}{C_{P}}}$ Ratio of potential difference in parallel to series combination, $\frac{\mathrm{V}_{\mathrm{P}}}{\mathrm{V}_{\mathrm{S}}}=\sqrt{\frac{\mathrm{C}}{3 \times 3 \mathrm{C}}}=\frac{1}{3}$
165819
The equivalent capacitance between the points $A$ and $B$ in the following circuit is :
1 $1 \mu \mathrm{F}$
2 $2 \mu \mathrm{F}$
3 $4 \mu \mathrm{F}$
4 $8 \mu \mathrm{F}$
Explanation:
: Capacitor of capacitance $1.5 \mu \mathrm{F}$ and $1.5 \mu \mathrm{F}$ are connected in parallel. Hence their equivalent resistance is $3 \mu \mathrm{F}$. Now, three capacitor each of value $3 \mu \mathrm{F}$ are in series. Hence, their equivalent capacitance is given by, $\frac{1}{\mathrm{C}}=\frac{1}{3}+\frac{1}{3}+\frac{1}{3}$ $\mathrm{C}=1 \mu \mathrm{F}$
[BCECE-2005
Capacitance
165820
When two conductors of charges and potentials $C_{1}, V_{1}$ and $C_{2}, V_{2}$ respectively are joined, the common potential will be:
: When two conductors of charges and potentials $\mathrm{C}_{1}, \mathrm{~V}_{1}$ and $\mathrm{C}_{2}, \mathrm{~V}_{2}$ respectively are joined. Let the common potential will be $\mathrm{V}$. By the law of conservation of charge, $\mathrm{Q}_{\mathrm{i}}=\mathrm{C}_{1} \mathrm{~V}_{1}+\mathrm{C}_{2} \mathrm{~V}_{2}$ $\mathrm{Q}_{\mathrm{f}}=\mathrm{C}_{1} \mathrm{~V}+\mathrm{C}_{2} \mathrm{~V}$ $\mathrm{Q}_{\mathrm{i}}=\mathrm{Q}_{\mathrm{f}}$ $\mathrm{C}_{1} \mathrm{~V}_{1}+\mathrm{C}_{2} \mathrm{~V}_{2}=\mathrm{C}_{1} \mathrm{~V}+\mathrm{C}_{2} \mathrm{~V}$ $\mathrm{~V}=\frac{\mathrm{C}_{1} \mathrm{~V}_{1}+\mathrm{C}_{2} \mathrm{~V}_{2}}{\mathrm{C}_{1}+\mathrm{C}_{2}}$
[BCECE-2003]
Capacitance
165821
In the circuit shown, the potential difference across the $4.5 \mu \mathrm{F}$ capacitor is
1 6 volt
2 $\frac{8}{3}$ volt
3 4 volt
4 8 volt
Explanation:
: Since, $4.6 \mu \mathrm{F}$ and $\mathrm{C}_{\mathrm{R}}=9 \mu \mathrm{F}$ are in series. So, the charge on capacitors is same. $\therefore \mathrm{C}_{1} \mathrm{~V}_{1}=\mathrm{C}_{2} \mathrm{~V}_{2}$ $\Rightarrow 9 \mathrm{~V}_{1}=4.5 \mathrm{~V}_{2}$ $\Rightarrow \mathrm{V}_{2}=2 \mathrm{~V}_{1}$ Given, $V_{1}+V_{2}=12$ Volt $\mathrm{V}_{1}+2 \mathrm{~V}_{1}=12$ $\Rightarrow 3 \mathrm{~V}_{1}=12$ $\Rightarrow \mathrm{V}_{1}=4$ volt $\mathrm{V}_{2}($ Voltage on $4 \mu \mathrm{F})=2 \times 4=8 \mathrm{~V}$
[MHT-CET 2020
Capacitance
165822
The potential differences that must be applied across the parallel and series combination of 3 identical capacitors are such that the energy stored in them becomes the same. The ratio of potential difference in parallel to series combination is
1 $\frac{1}{3}$
2 $\frac{1}{4}$
3 $\frac{1}{8}$
4 $\frac{1}{6}$
Explanation:
: Three capacitors are connected in parallel $\mathrm{C}_{\mathrm{P}}=\mathrm{C}+\mathrm{C}+\mathrm{C}$ $\mathrm{C}_{\mathrm{P}}=3 \mathrm{C}$ Three capacitors are connected in series, $\frac{1}{\mathrm{C}_{\mathrm{S}}}=\frac{1}{\mathrm{C}}+\frac{1}{\mathrm{C}}+\frac{1}{\mathrm{C}}$ $\mathrm{C}_{\mathrm{S}}=\frac{\mathrm{C}}{3}$ Energy stored is same in both conditions, $\frac{1}{2} C_{P} V_{P}^{2}=\frac{1}{2} C_{S} V_{S}^{2}$ $\left(\frac{V_{P}}{V_{S}}\right)^{2}=\frac{C_{S}}{C_{P}}$ $\frac{V_{P}}{V_{S}}=\sqrt{\frac{C_{S}}{C_{P}}}$ Ratio of potential difference in parallel to series combination, $\frac{\mathrm{V}_{\mathrm{P}}}{\mathrm{V}_{\mathrm{S}}}=\sqrt{\frac{\mathrm{C}}{3 \times 3 \mathrm{C}}}=\frac{1}{3}$
