165837
Four capacitors are connected as shown in the figure below. The equivalent capacitance between the points $P$ and $Q$ is
1 $4 \mu \mathrm{F}$
2 $\frac{1}{4} \mu \mathrm{F}$
3 $\frac{3}{4} \mu \mathrm{F}$
4 $\frac{4}{3} \mu \mathrm{F}$
Explanation:
: Given figure, Three, $1 \mu \mathrm{F}$ capacitors are connected in series combination. So, their resultant capacitance, $\frac{1}{\mathrm{C}_{1}}=\frac{1}{1}+\frac{1}{1}+\frac{1}{1}=3$ $\mathrm{C}_{1}=\frac{1}{3} \mu \mathrm{F}$ Now, capacitor $C_{1}$ is connected in parallel combination with $1 \mu \mathrm{F}$ capacitor. Hence, equivalent capacitance between the $\mathrm{P}$ and $\mathrm{Q}$, $\mathrm{C}_{\mathrm{PQ}}=\mathrm{C}_{1}+1$ $\mathrm{C}_{\mathrm{PQ}}=\frac{1}{3}+1=\frac{4}{3} \mu \mathrm{F}$
[CG PET -2018]
Capacitance
165838
Three capacitors each of $4 \mu \mathrm{F}$ are to be connected in such a way that the effective capacitance is $6 \mu \mathrm{F}$. This can be done by
1 Connecting all of them in series
2 Connecting them in parallel
3 Connecting two in series and one in parallel
4 Connecting two in parallel and one in series
Explanation:
: To get the equivalent capacitance $6 \mu \mathrm{F}$, the capacitor are arranged in the following manner. $\mathrm{C}_{1}$ and $\mathrm{C}_{2}$ are in series combination- $\frac{1}{\mathrm{C}_{\text {eq }}}=\frac{1}{\mathrm{C}_{1}}+\frac{1}{\mathrm{C}_{2}}$ $\frac{1}{\mathrm{C}_{\mathrm{eq}}}=\frac{1}{4}+\frac{1}{4}=\frac{1+1}{4}=\frac{2}{4}$ $\frac{1}{\mathrm{C}_{\mathrm{eq}}}=\frac{1}{2}$ $\mathrm{C}_{\mathrm{eq}}=2 \mu \mathrm{F}$ $\therefore \mathrm{C}_{\mathrm{eq}}$ and $\mathrm{C}_{3}$ capacitor are joined into the parallel combination- $\mathrm{C}_{\mathrm{eq}}^{\prime}=\mathrm{C}_{\mathrm{eq}}+\mathrm{C}_{3}$ $\mathrm{C}_{\mathrm{eq}}^{\prime}=2+4$ $\mathrm{C}_{\mathrm{eq}}^{\prime}=6 \mu \mathrm{F}$
[CG PET- 2017]
Capacitance
165839
Three capacitors $C_{1}, C_{2}$ and $C_{3}$ are connected to a battery of $\mathrm{V}$ volt as shown in figure. The charges and potentials are shown in figure. Then, the correct answer is
: Given circuit, According to Kirchoff's first law- We know that in series combination charge remains same and in parallel combination of capacitors voltage remain same. $\mathrm{Q}_{1}=\mathrm{Q}_{2}+\mathrm{Q}_{3}$ As $\mathrm{C}_{2}$ and $\mathrm{C}_{3}$ are in parallel, Then, $\quad \mathrm{V}_{2}=\mathrm{V}_{3}$ Total voltage $\mathrm{V}$ will equal to the combination or voltage across $\mathrm{C}_{1}$ and voltage across parallel combination of $\mathrm{C}_{2}$ and $\mathrm{C}_{3}$. So, $\mathrm{V}=\mathrm{V}_{1}+\mathrm{V}_{2} \text { and } \mathrm{Q}_{1}=\mathrm{Q}_{2}+\mathrm{Q}_{3}$
[CG PET -2016]
Capacitance
165840
Equivalent capacitance between $A$ and $B$ for circuit shown in figure.
1 $3 \mu \mathrm{F}$
2 $2 \mu \mathrm{F}$
3 $4 \mu \mathrm{F}$
4 $8 \mu \mathrm{F}$
Explanation:
:Given that, Hence, net capacitance between A \& B $\mathrm{C}_{\text {net }}=\mathrm{C}_{1}+\mathrm{C}_{2}$ $=2+2$ $\mathrm{C}_{\text {net }}=4 \mu \mathrm{F}$
[CG PET- 2007]
Capacitance
165841
Potential difference across capacitor $4.5 \mu \mathrm{F}$ capacitance is
1 $\frac{8}{3} \mathrm{~V}$
2 $4 \mathrm{~V}$
3 $6 \mathrm{~V}$
4 $8 \mathrm{~V}$
Explanation:
: Equivalent capacitance is given by- We know that, $\mathrm{Q}=\mathrm{C} \cdot \mathrm{V}$ $\mathrm{V}=\frac{\mathrm{Q}}{\mathrm{C}}$ $\mathrm{V} \propto \frac{1}{\mathrm{C}}$ Hence, $\mathrm{V}_{1}: \mathrm{V}_{2}=\mathrm{C}_{2}: \mathrm{C}_{1}$ $\mathrm{V}_{1} =\frac{\mathrm{C}_{2}}{\mathrm{C}_{1}+\mathrm{C}_{2}} \times \mathrm{V}$ $\mathrm{V}_{1} =\frac{9}{13.5} \times 12$ $\mathrm{~V}_{1} =\frac{90}{135} \times 12$ $\mathrm{~V}_{1} =8 \text { Volt }$
165837
Four capacitors are connected as shown in the figure below. The equivalent capacitance between the points $P$ and $Q$ is
1 $4 \mu \mathrm{F}$
2 $\frac{1}{4} \mu \mathrm{F}$
3 $\frac{3}{4} \mu \mathrm{F}$
4 $\frac{4}{3} \mu \mathrm{F}$
Explanation:
: Given figure, Three, $1 \mu \mathrm{F}$ capacitors are connected in series combination. So, their resultant capacitance, $\frac{1}{\mathrm{C}_{1}}=\frac{1}{1}+\frac{1}{1}+\frac{1}{1}=3$ $\mathrm{C}_{1}=\frac{1}{3} \mu \mathrm{F}$ Now, capacitor $C_{1}$ is connected in parallel combination with $1 \mu \mathrm{F}$ capacitor. Hence, equivalent capacitance between the $\mathrm{P}$ and $\mathrm{Q}$, $\mathrm{C}_{\mathrm{PQ}}=\mathrm{C}_{1}+1$ $\mathrm{C}_{\mathrm{PQ}}=\frac{1}{3}+1=\frac{4}{3} \mu \mathrm{F}$
[CG PET -2018]
Capacitance
165838
Three capacitors each of $4 \mu \mathrm{F}$ are to be connected in such a way that the effective capacitance is $6 \mu \mathrm{F}$. This can be done by
1 Connecting all of them in series
2 Connecting them in parallel
3 Connecting two in series and one in parallel
4 Connecting two in parallel and one in series
Explanation:
: To get the equivalent capacitance $6 \mu \mathrm{F}$, the capacitor are arranged in the following manner. $\mathrm{C}_{1}$ and $\mathrm{C}_{2}$ are in series combination- $\frac{1}{\mathrm{C}_{\text {eq }}}=\frac{1}{\mathrm{C}_{1}}+\frac{1}{\mathrm{C}_{2}}$ $\frac{1}{\mathrm{C}_{\mathrm{eq}}}=\frac{1}{4}+\frac{1}{4}=\frac{1+1}{4}=\frac{2}{4}$ $\frac{1}{\mathrm{C}_{\mathrm{eq}}}=\frac{1}{2}$ $\mathrm{C}_{\mathrm{eq}}=2 \mu \mathrm{F}$ $\therefore \mathrm{C}_{\mathrm{eq}}$ and $\mathrm{C}_{3}$ capacitor are joined into the parallel combination- $\mathrm{C}_{\mathrm{eq}}^{\prime}=\mathrm{C}_{\mathrm{eq}}+\mathrm{C}_{3}$ $\mathrm{C}_{\mathrm{eq}}^{\prime}=2+4$ $\mathrm{C}_{\mathrm{eq}}^{\prime}=6 \mu \mathrm{F}$
[CG PET- 2017]
Capacitance
165839
Three capacitors $C_{1}, C_{2}$ and $C_{3}$ are connected to a battery of $\mathrm{V}$ volt as shown in figure. The charges and potentials are shown in figure. Then, the correct answer is
: Given circuit, According to Kirchoff's first law- We know that in series combination charge remains same and in parallel combination of capacitors voltage remain same. $\mathrm{Q}_{1}=\mathrm{Q}_{2}+\mathrm{Q}_{3}$ As $\mathrm{C}_{2}$ and $\mathrm{C}_{3}$ are in parallel, Then, $\quad \mathrm{V}_{2}=\mathrm{V}_{3}$ Total voltage $\mathrm{V}$ will equal to the combination or voltage across $\mathrm{C}_{1}$ and voltage across parallel combination of $\mathrm{C}_{2}$ and $\mathrm{C}_{3}$. So, $\mathrm{V}=\mathrm{V}_{1}+\mathrm{V}_{2} \text { and } \mathrm{Q}_{1}=\mathrm{Q}_{2}+\mathrm{Q}_{3}$
[CG PET -2016]
Capacitance
165840
Equivalent capacitance between $A$ and $B$ for circuit shown in figure.
1 $3 \mu \mathrm{F}$
2 $2 \mu \mathrm{F}$
3 $4 \mu \mathrm{F}$
4 $8 \mu \mathrm{F}$
Explanation:
:Given that, Hence, net capacitance between A \& B $\mathrm{C}_{\text {net }}=\mathrm{C}_{1}+\mathrm{C}_{2}$ $=2+2$ $\mathrm{C}_{\text {net }}=4 \mu \mathrm{F}$
[CG PET- 2007]
Capacitance
165841
Potential difference across capacitor $4.5 \mu \mathrm{F}$ capacitance is
1 $\frac{8}{3} \mathrm{~V}$
2 $4 \mathrm{~V}$
3 $6 \mathrm{~V}$
4 $8 \mathrm{~V}$
Explanation:
: Equivalent capacitance is given by- We know that, $\mathrm{Q}=\mathrm{C} \cdot \mathrm{V}$ $\mathrm{V}=\frac{\mathrm{Q}}{\mathrm{C}}$ $\mathrm{V} \propto \frac{1}{\mathrm{C}}$ Hence, $\mathrm{V}_{1}: \mathrm{V}_{2}=\mathrm{C}_{2}: \mathrm{C}_{1}$ $\mathrm{V}_{1} =\frac{\mathrm{C}_{2}}{\mathrm{C}_{1}+\mathrm{C}_{2}} \times \mathrm{V}$ $\mathrm{V}_{1} =\frac{9}{13.5} \times 12$ $\mathrm{~V}_{1} =\frac{90}{135} \times 12$ $\mathrm{~V}_{1} =8 \text { Volt }$
NEET Test Series from KOTA - 10 Papers In MS WORD
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Capacitance
165837
Four capacitors are connected as shown in the figure below. The equivalent capacitance between the points $P$ and $Q$ is
1 $4 \mu \mathrm{F}$
2 $\frac{1}{4} \mu \mathrm{F}$
3 $\frac{3}{4} \mu \mathrm{F}$
4 $\frac{4}{3} \mu \mathrm{F}$
Explanation:
: Given figure, Three, $1 \mu \mathrm{F}$ capacitors are connected in series combination. So, their resultant capacitance, $\frac{1}{\mathrm{C}_{1}}=\frac{1}{1}+\frac{1}{1}+\frac{1}{1}=3$ $\mathrm{C}_{1}=\frac{1}{3} \mu \mathrm{F}$ Now, capacitor $C_{1}$ is connected in parallel combination with $1 \mu \mathrm{F}$ capacitor. Hence, equivalent capacitance between the $\mathrm{P}$ and $\mathrm{Q}$, $\mathrm{C}_{\mathrm{PQ}}=\mathrm{C}_{1}+1$ $\mathrm{C}_{\mathrm{PQ}}=\frac{1}{3}+1=\frac{4}{3} \mu \mathrm{F}$
[CG PET -2018]
Capacitance
165838
Three capacitors each of $4 \mu \mathrm{F}$ are to be connected in such a way that the effective capacitance is $6 \mu \mathrm{F}$. This can be done by
1 Connecting all of them in series
2 Connecting them in parallel
3 Connecting two in series and one in parallel
4 Connecting two in parallel and one in series
Explanation:
: To get the equivalent capacitance $6 \mu \mathrm{F}$, the capacitor are arranged in the following manner. $\mathrm{C}_{1}$ and $\mathrm{C}_{2}$ are in series combination- $\frac{1}{\mathrm{C}_{\text {eq }}}=\frac{1}{\mathrm{C}_{1}}+\frac{1}{\mathrm{C}_{2}}$ $\frac{1}{\mathrm{C}_{\mathrm{eq}}}=\frac{1}{4}+\frac{1}{4}=\frac{1+1}{4}=\frac{2}{4}$ $\frac{1}{\mathrm{C}_{\mathrm{eq}}}=\frac{1}{2}$ $\mathrm{C}_{\mathrm{eq}}=2 \mu \mathrm{F}$ $\therefore \mathrm{C}_{\mathrm{eq}}$ and $\mathrm{C}_{3}$ capacitor are joined into the parallel combination- $\mathrm{C}_{\mathrm{eq}}^{\prime}=\mathrm{C}_{\mathrm{eq}}+\mathrm{C}_{3}$ $\mathrm{C}_{\mathrm{eq}}^{\prime}=2+4$ $\mathrm{C}_{\mathrm{eq}}^{\prime}=6 \mu \mathrm{F}$
[CG PET- 2017]
Capacitance
165839
Three capacitors $C_{1}, C_{2}$ and $C_{3}$ are connected to a battery of $\mathrm{V}$ volt as shown in figure. The charges and potentials are shown in figure. Then, the correct answer is
: Given circuit, According to Kirchoff's first law- We know that in series combination charge remains same and in parallel combination of capacitors voltage remain same. $\mathrm{Q}_{1}=\mathrm{Q}_{2}+\mathrm{Q}_{3}$ As $\mathrm{C}_{2}$ and $\mathrm{C}_{3}$ are in parallel, Then, $\quad \mathrm{V}_{2}=\mathrm{V}_{3}$ Total voltage $\mathrm{V}$ will equal to the combination or voltage across $\mathrm{C}_{1}$ and voltage across parallel combination of $\mathrm{C}_{2}$ and $\mathrm{C}_{3}$. So, $\mathrm{V}=\mathrm{V}_{1}+\mathrm{V}_{2} \text { and } \mathrm{Q}_{1}=\mathrm{Q}_{2}+\mathrm{Q}_{3}$
[CG PET -2016]
Capacitance
165840
Equivalent capacitance between $A$ and $B$ for circuit shown in figure.
1 $3 \mu \mathrm{F}$
2 $2 \mu \mathrm{F}$
3 $4 \mu \mathrm{F}$
4 $8 \mu \mathrm{F}$
Explanation:
:Given that, Hence, net capacitance between A \& B $\mathrm{C}_{\text {net }}=\mathrm{C}_{1}+\mathrm{C}_{2}$ $=2+2$ $\mathrm{C}_{\text {net }}=4 \mu \mathrm{F}$
[CG PET- 2007]
Capacitance
165841
Potential difference across capacitor $4.5 \mu \mathrm{F}$ capacitance is
1 $\frac{8}{3} \mathrm{~V}$
2 $4 \mathrm{~V}$
3 $6 \mathrm{~V}$
4 $8 \mathrm{~V}$
Explanation:
: Equivalent capacitance is given by- We know that, $\mathrm{Q}=\mathrm{C} \cdot \mathrm{V}$ $\mathrm{V}=\frac{\mathrm{Q}}{\mathrm{C}}$ $\mathrm{V} \propto \frac{1}{\mathrm{C}}$ Hence, $\mathrm{V}_{1}: \mathrm{V}_{2}=\mathrm{C}_{2}: \mathrm{C}_{1}$ $\mathrm{V}_{1} =\frac{\mathrm{C}_{2}}{\mathrm{C}_{1}+\mathrm{C}_{2}} \times \mathrm{V}$ $\mathrm{V}_{1} =\frac{9}{13.5} \times 12$ $\mathrm{~V}_{1} =\frac{90}{135} \times 12$ $\mathrm{~V}_{1} =8 \text { Volt }$
165837
Four capacitors are connected as shown in the figure below. The equivalent capacitance between the points $P$ and $Q$ is
1 $4 \mu \mathrm{F}$
2 $\frac{1}{4} \mu \mathrm{F}$
3 $\frac{3}{4} \mu \mathrm{F}$
4 $\frac{4}{3} \mu \mathrm{F}$
Explanation:
: Given figure, Three, $1 \mu \mathrm{F}$ capacitors are connected in series combination. So, their resultant capacitance, $\frac{1}{\mathrm{C}_{1}}=\frac{1}{1}+\frac{1}{1}+\frac{1}{1}=3$ $\mathrm{C}_{1}=\frac{1}{3} \mu \mathrm{F}$ Now, capacitor $C_{1}$ is connected in parallel combination with $1 \mu \mathrm{F}$ capacitor. Hence, equivalent capacitance between the $\mathrm{P}$ and $\mathrm{Q}$, $\mathrm{C}_{\mathrm{PQ}}=\mathrm{C}_{1}+1$ $\mathrm{C}_{\mathrm{PQ}}=\frac{1}{3}+1=\frac{4}{3} \mu \mathrm{F}$
[CG PET -2018]
Capacitance
165838
Three capacitors each of $4 \mu \mathrm{F}$ are to be connected in such a way that the effective capacitance is $6 \mu \mathrm{F}$. This can be done by
1 Connecting all of them in series
2 Connecting them in parallel
3 Connecting two in series and one in parallel
4 Connecting two in parallel and one in series
Explanation:
: To get the equivalent capacitance $6 \mu \mathrm{F}$, the capacitor are arranged in the following manner. $\mathrm{C}_{1}$ and $\mathrm{C}_{2}$ are in series combination- $\frac{1}{\mathrm{C}_{\text {eq }}}=\frac{1}{\mathrm{C}_{1}}+\frac{1}{\mathrm{C}_{2}}$ $\frac{1}{\mathrm{C}_{\mathrm{eq}}}=\frac{1}{4}+\frac{1}{4}=\frac{1+1}{4}=\frac{2}{4}$ $\frac{1}{\mathrm{C}_{\mathrm{eq}}}=\frac{1}{2}$ $\mathrm{C}_{\mathrm{eq}}=2 \mu \mathrm{F}$ $\therefore \mathrm{C}_{\mathrm{eq}}$ and $\mathrm{C}_{3}$ capacitor are joined into the parallel combination- $\mathrm{C}_{\mathrm{eq}}^{\prime}=\mathrm{C}_{\mathrm{eq}}+\mathrm{C}_{3}$ $\mathrm{C}_{\mathrm{eq}}^{\prime}=2+4$ $\mathrm{C}_{\mathrm{eq}}^{\prime}=6 \mu \mathrm{F}$
[CG PET- 2017]
Capacitance
165839
Three capacitors $C_{1}, C_{2}$ and $C_{3}$ are connected to a battery of $\mathrm{V}$ volt as shown in figure. The charges and potentials are shown in figure. Then, the correct answer is
: Given circuit, According to Kirchoff's first law- We know that in series combination charge remains same and in parallel combination of capacitors voltage remain same. $\mathrm{Q}_{1}=\mathrm{Q}_{2}+\mathrm{Q}_{3}$ As $\mathrm{C}_{2}$ and $\mathrm{C}_{3}$ are in parallel, Then, $\quad \mathrm{V}_{2}=\mathrm{V}_{3}$ Total voltage $\mathrm{V}$ will equal to the combination or voltage across $\mathrm{C}_{1}$ and voltage across parallel combination of $\mathrm{C}_{2}$ and $\mathrm{C}_{3}$. So, $\mathrm{V}=\mathrm{V}_{1}+\mathrm{V}_{2} \text { and } \mathrm{Q}_{1}=\mathrm{Q}_{2}+\mathrm{Q}_{3}$
[CG PET -2016]
Capacitance
165840
Equivalent capacitance between $A$ and $B$ for circuit shown in figure.
1 $3 \mu \mathrm{F}$
2 $2 \mu \mathrm{F}$
3 $4 \mu \mathrm{F}$
4 $8 \mu \mathrm{F}$
Explanation:
:Given that, Hence, net capacitance between A \& B $\mathrm{C}_{\text {net }}=\mathrm{C}_{1}+\mathrm{C}_{2}$ $=2+2$ $\mathrm{C}_{\text {net }}=4 \mu \mathrm{F}$
[CG PET- 2007]
Capacitance
165841
Potential difference across capacitor $4.5 \mu \mathrm{F}$ capacitance is
1 $\frac{8}{3} \mathrm{~V}$
2 $4 \mathrm{~V}$
3 $6 \mathrm{~V}$
4 $8 \mathrm{~V}$
Explanation:
: Equivalent capacitance is given by- We know that, $\mathrm{Q}=\mathrm{C} \cdot \mathrm{V}$ $\mathrm{V}=\frac{\mathrm{Q}}{\mathrm{C}}$ $\mathrm{V} \propto \frac{1}{\mathrm{C}}$ Hence, $\mathrm{V}_{1}: \mathrm{V}_{2}=\mathrm{C}_{2}: \mathrm{C}_{1}$ $\mathrm{V}_{1} =\frac{\mathrm{C}_{2}}{\mathrm{C}_{1}+\mathrm{C}_{2}} \times \mathrm{V}$ $\mathrm{V}_{1} =\frac{9}{13.5} \times 12$ $\mathrm{~V}_{1} =\frac{90}{135} \times 12$ $\mathrm{~V}_{1} =8 \text { Volt }$
165837
Four capacitors are connected as shown in the figure below. The equivalent capacitance between the points $P$ and $Q$ is
1 $4 \mu \mathrm{F}$
2 $\frac{1}{4} \mu \mathrm{F}$
3 $\frac{3}{4} \mu \mathrm{F}$
4 $\frac{4}{3} \mu \mathrm{F}$
Explanation:
: Given figure, Three, $1 \mu \mathrm{F}$ capacitors are connected in series combination. So, their resultant capacitance, $\frac{1}{\mathrm{C}_{1}}=\frac{1}{1}+\frac{1}{1}+\frac{1}{1}=3$ $\mathrm{C}_{1}=\frac{1}{3} \mu \mathrm{F}$ Now, capacitor $C_{1}$ is connected in parallel combination with $1 \mu \mathrm{F}$ capacitor. Hence, equivalent capacitance between the $\mathrm{P}$ and $\mathrm{Q}$, $\mathrm{C}_{\mathrm{PQ}}=\mathrm{C}_{1}+1$ $\mathrm{C}_{\mathrm{PQ}}=\frac{1}{3}+1=\frac{4}{3} \mu \mathrm{F}$
[CG PET -2018]
Capacitance
165838
Three capacitors each of $4 \mu \mathrm{F}$ are to be connected in such a way that the effective capacitance is $6 \mu \mathrm{F}$. This can be done by
1 Connecting all of them in series
2 Connecting them in parallel
3 Connecting two in series and one in parallel
4 Connecting two in parallel and one in series
Explanation:
: To get the equivalent capacitance $6 \mu \mathrm{F}$, the capacitor are arranged in the following manner. $\mathrm{C}_{1}$ and $\mathrm{C}_{2}$ are in series combination- $\frac{1}{\mathrm{C}_{\text {eq }}}=\frac{1}{\mathrm{C}_{1}}+\frac{1}{\mathrm{C}_{2}}$ $\frac{1}{\mathrm{C}_{\mathrm{eq}}}=\frac{1}{4}+\frac{1}{4}=\frac{1+1}{4}=\frac{2}{4}$ $\frac{1}{\mathrm{C}_{\mathrm{eq}}}=\frac{1}{2}$ $\mathrm{C}_{\mathrm{eq}}=2 \mu \mathrm{F}$ $\therefore \mathrm{C}_{\mathrm{eq}}$ and $\mathrm{C}_{3}$ capacitor are joined into the parallel combination- $\mathrm{C}_{\mathrm{eq}}^{\prime}=\mathrm{C}_{\mathrm{eq}}+\mathrm{C}_{3}$ $\mathrm{C}_{\mathrm{eq}}^{\prime}=2+4$ $\mathrm{C}_{\mathrm{eq}}^{\prime}=6 \mu \mathrm{F}$
[CG PET- 2017]
Capacitance
165839
Three capacitors $C_{1}, C_{2}$ and $C_{3}$ are connected to a battery of $\mathrm{V}$ volt as shown in figure. The charges and potentials are shown in figure. Then, the correct answer is
: Given circuit, According to Kirchoff's first law- We know that in series combination charge remains same and in parallel combination of capacitors voltage remain same. $\mathrm{Q}_{1}=\mathrm{Q}_{2}+\mathrm{Q}_{3}$ As $\mathrm{C}_{2}$ and $\mathrm{C}_{3}$ are in parallel, Then, $\quad \mathrm{V}_{2}=\mathrm{V}_{3}$ Total voltage $\mathrm{V}$ will equal to the combination or voltage across $\mathrm{C}_{1}$ and voltage across parallel combination of $\mathrm{C}_{2}$ and $\mathrm{C}_{3}$. So, $\mathrm{V}=\mathrm{V}_{1}+\mathrm{V}_{2} \text { and } \mathrm{Q}_{1}=\mathrm{Q}_{2}+\mathrm{Q}_{3}$
[CG PET -2016]
Capacitance
165840
Equivalent capacitance between $A$ and $B$ for circuit shown in figure.
1 $3 \mu \mathrm{F}$
2 $2 \mu \mathrm{F}$
3 $4 \mu \mathrm{F}$
4 $8 \mu \mathrm{F}$
Explanation:
:Given that, Hence, net capacitance between A \& B $\mathrm{C}_{\text {net }}=\mathrm{C}_{1}+\mathrm{C}_{2}$ $=2+2$ $\mathrm{C}_{\text {net }}=4 \mu \mathrm{F}$
[CG PET- 2007]
Capacitance
165841
Potential difference across capacitor $4.5 \mu \mathrm{F}$ capacitance is
1 $\frac{8}{3} \mathrm{~V}$
2 $4 \mathrm{~V}$
3 $6 \mathrm{~V}$
4 $8 \mathrm{~V}$
Explanation:
: Equivalent capacitance is given by- We know that, $\mathrm{Q}=\mathrm{C} \cdot \mathrm{V}$ $\mathrm{V}=\frac{\mathrm{Q}}{\mathrm{C}}$ $\mathrm{V} \propto \frac{1}{\mathrm{C}}$ Hence, $\mathrm{V}_{1}: \mathrm{V}_{2}=\mathrm{C}_{2}: \mathrm{C}_{1}$ $\mathrm{V}_{1} =\frac{\mathrm{C}_{2}}{\mathrm{C}_{1}+\mathrm{C}_{2}} \times \mathrm{V}$ $\mathrm{V}_{1} =\frac{9}{13.5} \times 12$ $\mathrm{~V}_{1} =\frac{90}{135} \times 12$ $\mathrm{~V}_{1} =8 \text { Volt }$
