165796
An infinite number of capacitors each of capacity $2 \mu \mathrm{F}$ are connected as shown in the figure. The resultant capacity between $A$ and $B$ is
1 $5 \mu \mathrm{F}$
2 $1 \mu \mathrm{F}$
3 $3 \mu \mathrm{F}$
4 $4 \mu \mathrm{F}$
Explanation:
: Equivalent capacitance of Capacitors which are in series, $\frac{1}{\mathrm{C}_{\text {eq }}}=\frac{1}{\mathrm{C}_{1}}+\frac{1}{\mathrm{C}_{2}}+\ldots \ldots \ldots .+\frac{1}{\mathrm{C}_{\mathrm{n}}}$ For $\mathrm{C}_{1}=\mathrm{C}_{2}=\ldots \ldots \ldots \mathrm{C}_{\mathrm{n}}=2 \mu \mathrm{F}$ $\mathrm{C}_{\mathrm{eq}}=\frac{\mathrm{C}}{\mathrm{n}}$ For first row $\mathrm{C}_{\mathrm{A}}=\mathrm{C}$ For second row $\mathrm{C}_{\mathrm{B}}=\frac{\mathrm{C}}{2}$ For third row $\mathrm{C}_{\mathrm{C}}=\frac{\mathrm{C}}{4}$ and so on . From figure all the capacitor are in parallel. $\mathrm{C}_{\text {final }}=\mathrm{C}_{\mathrm{A}}+\mathrm{C}_{\mathrm{B}}+\mathrm{C}_{\mathrm{C}}+\ldots \ldots \ldots+\mathrm{C}_{\infty}$ $\mathrm{C}_{\text {final }}=\mathrm{C}+\frac{\mathrm{C}}{2}+\frac{\mathrm{C}}{4}+\frac{\mathrm{C}}{8}+\ldots \ldots \ldots+\frac{\mathrm{C}}{\infty}$ $\mathrm{C}_{\text {final }}=\frac{\mathrm{C}}{1-\frac{1}{2}}=2 \mathrm{C}$ $\mathrm{C}_{\text {final }}=2 \times 2 \times 10^{-6} \quad(\because \mathrm{C}=2 \mu \mathrm{F})$ $\mathrm{C}_{\text {final }}=4 \mu \mathrm{F}$
[AP EAMCET-28.04.2017
Capacitance
165797
Three capacitors each of capacitance $\mathbf{C}$ and of breakdown voltage $V$ are joined in series. The capacitance and breakdown voltage of the combination will be
1 $\frac{\mathrm{C}}{3}, \frac{\mathrm{V}}{3}$
2 $3 \mathrm{C}, \frac{\mathrm{V}}{3}$
3 $\frac{\mathrm{C}}{3}, 3 \mathrm{~V}$
4 $3 \mathrm{C}, 3 \mathrm{~V}$
Explanation:
: In series, combination of capacitors $\mathrm{V}_{\text {eff }}=\mathrm{V}+\mathrm{V}+\mathrm{V}=3 \mathrm{~V}$ And, $\frac{1}{\mathrm{C}_{\mathrm{eff}}}=\frac{1}{\mathrm{C}}+\frac{1}{\mathrm{C}}+\frac{1}{\mathrm{C}}$ $\frac{1}{C_{\text {eff }}}=\frac{3}{C}$ $C_{\text {eff }}=\frac{C}{3}$ Thus, the capacitance and break down voltage of the combination will be $\mathrm{C} / 3$ and $3 \mathrm{~V}$.
[AIPMT-2009]
Capacitance
165798
The equivalent capacitance of the combination shown in the figure is
1 $3 \mathrm{C}$
2 $2 \mathrm{C}$
3 $\frac{\mathrm{C}}{2}$
4 $\frac{3 \mathrm{C}}{2}$
Explanation:
: In this circuit one capacitor is shorted. So, the remaining two capacitor are in parallel. Given, $\mathrm{C}_{1}=\mathrm{C}_{2}=\mathrm{C}$ So, the equivalent capacitance is - $\mathrm{C}_{\mathrm{eq}}=\mathrm{C}_{1}+\mathrm{C}_{2}=2 \mathrm{C}$
[NEET-2021]
Capacitance
165799
A parallel plate capacitor has 91 plates, all are identical and arranged with same spacing between them. If the capacitance between adjacent plates is $3 \mathrm{pF}$. What will be the resultant capacitance?
1 $273 \mathrm{pF}$
2 $30 \mathrm{pF}$
3 $94 \mathrm{pF}$
4 $270 \mathrm{pF}$
Explanation:
: Given, $\mathrm{n}=91, \mathrm{C}=3 \mathrm{pF}$ When $\mathrm{n}$ plates are arranged with the same spacing between them, the combination behaves as $(n-1)$ capacitor in parallel since charge on each plate is same. So, equivalent capacitance is $\mathrm{C}_{\mathrm{eq}}=\mathrm{C}(\mathrm{n}-1)$ $=3(91-1)$ $=270 \mathrm{pF}$
NEET Test Series from KOTA - 10 Papers In MS WORD
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Capacitance
165796
An infinite number of capacitors each of capacity $2 \mu \mathrm{F}$ are connected as shown in the figure. The resultant capacity between $A$ and $B$ is
1 $5 \mu \mathrm{F}$
2 $1 \mu \mathrm{F}$
3 $3 \mu \mathrm{F}$
4 $4 \mu \mathrm{F}$
Explanation:
: Equivalent capacitance of Capacitors which are in series, $\frac{1}{\mathrm{C}_{\text {eq }}}=\frac{1}{\mathrm{C}_{1}}+\frac{1}{\mathrm{C}_{2}}+\ldots \ldots \ldots .+\frac{1}{\mathrm{C}_{\mathrm{n}}}$ For $\mathrm{C}_{1}=\mathrm{C}_{2}=\ldots \ldots \ldots \mathrm{C}_{\mathrm{n}}=2 \mu \mathrm{F}$ $\mathrm{C}_{\mathrm{eq}}=\frac{\mathrm{C}}{\mathrm{n}}$ For first row $\mathrm{C}_{\mathrm{A}}=\mathrm{C}$ For second row $\mathrm{C}_{\mathrm{B}}=\frac{\mathrm{C}}{2}$ For third row $\mathrm{C}_{\mathrm{C}}=\frac{\mathrm{C}}{4}$ and so on . From figure all the capacitor are in parallel. $\mathrm{C}_{\text {final }}=\mathrm{C}_{\mathrm{A}}+\mathrm{C}_{\mathrm{B}}+\mathrm{C}_{\mathrm{C}}+\ldots \ldots \ldots+\mathrm{C}_{\infty}$ $\mathrm{C}_{\text {final }}=\mathrm{C}+\frac{\mathrm{C}}{2}+\frac{\mathrm{C}}{4}+\frac{\mathrm{C}}{8}+\ldots \ldots \ldots+\frac{\mathrm{C}}{\infty}$ $\mathrm{C}_{\text {final }}=\frac{\mathrm{C}}{1-\frac{1}{2}}=2 \mathrm{C}$ $\mathrm{C}_{\text {final }}=2 \times 2 \times 10^{-6} \quad(\because \mathrm{C}=2 \mu \mathrm{F})$ $\mathrm{C}_{\text {final }}=4 \mu \mathrm{F}$
[AP EAMCET-28.04.2017
Capacitance
165797
Three capacitors each of capacitance $\mathbf{C}$ and of breakdown voltage $V$ are joined in series. The capacitance and breakdown voltage of the combination will be
1 $\frac{\mathrm{C}}{3}, \frac{\mathrm{V}}{3}$
2 $3 \mathrm{C}, \frac{\mathrm{V}}{3}$
3 $\frac{\mathrm{C}}{3}, 3 \mathrm{~V}$
4 $3 \mathrm{C}, 3 \mathrm{~V}$
Explanation:
: In series, combination of capacitors $\mathrm{V}_{\text {eff }}=\mathrm{V}+\mathrm{V}+\mathrm{V}=3 \mathrm{~V}$ And, $\frac{1}{\mathrm{C}_{\mathrm{eff}}}=\frac{1}{\mathrm{C}}+\frac{1}{\mathrm{C}}+\frac{1}{\mathrm{C}}$ $\frac{1}{C_{\text {eff }}}=\frac{3}{C}$ $C_{\text {eff }}=\frac{C}{3}$ Thus, the capacitance and break down voltage of the combination will be $\mathrm{C} / 3$ and $3 \mathrm{~V}$.
[AIPMT-2009]
Capacitance
165798
The equivalent capacitance of the combination shown in the figure is
1 $3 \mathrm{C}$
2 $2 \mathrm{C}$
3 $\frac{\mathrm{C}}{2}$
4 $\frac{3 \mathrm{C}}{2}$
Explanation:
: In this circuit one capacitor is shorted. So, the remaining two capacitor are in parallel. Given, $\mathrm{C}_{1}=\mathrm{C}_{2}=\mathrm{C}$ So, the equivalent capacitance is - $\mathrm{C}_{\mathrm{eq}}=\mathrm{C}_{1}+\mathrm{C}_{2}=2 \mathrm{C}$
[NEET-2021]
Capacitance
165799
A parallel plate capacitor has 91 plates, all are identical and arranged with same spacing between them. If the capacitance between adjacent plates is $3 \mathrm{pF}$. What will be the resultant capacitance?
1 $273 \mathrm{pF}$
2 $30 \mathrm{pF}$
3 $94 \mathrm{pF}$
4 $270 \mathrm{pF}$
Explanation:
: Given, $\mathrm{n}=91, \mathrm{C}=3 \mathrm{pF}$ When $\mathrm{n}$ plates are arranged with the same spacing between them, the combination behaves as $(n-1)$ capacitor in parallel since charge on each plate is same. So, equivalent capacitance is $\mathrm{C}_{\mathrm{eq}}=\mathrm{C}(\mathrm{n}-1)$ $=3(91-1)$ $=270 \mathrm{pF}$
165796
An infinite number of capacitors each of capacity $2 \mu \mathrm{F}$ are connected as shown in the figure. The resultant capacity between $A$ and $B$ is
1 $5 \mu \mathrm{F}$
2 $1 \mu \mathrm{F}$
3 $3 \mu \mathrm{F}$
4 $4 \mu \mathrm{F}$
Explanation:
: Equivalent capacitance of Capacitors which are in series, $\frac{1}{\mathrm{C}_{\text {eq }}}=\frac{1}{\mathrm{C}_{1}}+\frac{1}{\mathrm{C}_{2}}+\ldots \ldots \ldots .+\frac{1}{\mathrm{C}_{\mathrm{n}}}$ For $\mathrm{C}_{1}=\mathrm{C}_{2}=\ldots \ldots \ldots \mathrm{C}_{\mathrm{n}}=2 \mu \mathrm{F}$ $\mathrm{C}_{\mathrm{eq}}=\frac{\mathrm{C}}{\mathrm{n}}$ For first row $\mathrm{C}_{\mathrm{A}}=\mathrm{C}$ For second row $\mathrm{C}_{\mathrm{B}}=\frac{\mathrm{C}}{2}$ For third row $\mathrm{C}_{\mathrm{C}}=\frac{\mathrm{C}}{4}$ and so on . From figure all the capacitor are in parallel. $\mathrm{C}_{\text {final }}=\mathrm{C}_{\mathrm{A}}+\mathrm{C}_{\mathrm{B}}+\mathrm{C}_{\mathrm{C}}+\ldots \ldots \ldots+\mathrm{C}_{\infty}$ $\mathrm{C}_{\text {final }}=\mathrm{C}+\frac{\mathrm{C}}{2}+\frac{\mathrm{C}}{4}+\frac{\mathrm{C}}{8}+\ldots \ldots \ldots+\frac{\mathrm{C}}{\infty}$ $\mathrm{C}_{\text {final }}=\frac{\mathrm{C}}{1-\frac{1}{2}}=2 \mathrm{C}$ $\mathrm{C}_{\text {final }}=2 \times 2 \times 10^{-6} \quad(\because \mathrm{C}=2 \mu \mathrm{F})$ $\mathrm{C}_{\text {final }}=4 \mu \mathrm{F}$
[AP EAMCET-28.04.2017
Capacitance
165797
Three capacitors each of capacitance $\mathbf{C}$ and of breakdown voltage $V$ are joined in series. The capacitance and breakdown voltage of the combination will be
1 $\frac{\mathrm{C}}{3}, \frac{\mathrm{V}}{3}$
2 $3 \mathrm{C}, \frac{\mathrm{V}}{3}$
3 $\frac{\mathrm{C}}{3}, 3 \mathrm{~V}$
4 $3 \mathrm{C}, 3 \mathrm{~V}$
Explanation:
: In series, combination of capacitors $\mathrm{V}_{\text {eff }}=\mathrm{V}+\mathrm{V}+\mathrm{V}=3 \mathrm{~V}$ And, $\frac{1}{\mathrm{C}_{\mathrm{eff}}}=\frac{1}{\mathrm{C}}+\frac{1}{\mathrm{C}}+\frac{1}{\mathrm{C}}$ $\frac{1}{C_{\text {eff }}}=\frac{3}{C}$ $C_{\text {eff }}=\frac{C}{3}$ Thus, the capacitance and break down voltage of the combination will be $\mathrm{C} / 3$ and $3 \mathrm{~V}$.
[AIPMT-2009]
Capacitance
165798
The equivalent capacitance of the combination shown in the figure is
1 $3 \mathrm{C}$
2 $2 \mathrm{C}$
3 $\frac{\mathrm{C}}{2}$
4 $\frac{3 \mathrm{C}}{2}$
Explanation:
: In this circuit one capacitor is shorted. So, the remaining two capacitor are in parallel. Given, $\mathrm{C}_{1}=\mathrm{C}_{2}=\mathrm{C}$ So, the equivalent capacitance is - $\mathrm{C}_{\mathrm{eq}}=\mathrm{C}_{1}+\mathrm{C}_{2}=2 \mathrm{C}$
[NEET-2021]
Capacitance
165799
A parallel plate capacitor has 91 plates, all are identical and arranged with same spacing between them. If the capacitance between adjacent plates is $3 \mathrm{pF}$. What will be the resultant capacitance?
1 $273 \mathrm{pF}$
2 $30 \mathrm{pF}$
3 $94 \mathrm{pF}$
4 $270 \mathrm{pF}$
Explanation:
: Given, $\mathrm{n}=91, \mathrm{C}=3 \mathrm{pF}$ When $\mathrm{n}$ plates are arranged with the same spacing between them, the combination behaves as $(n-1)$ capacitor in parallel since charge on each plate is same. So, equivalent capacitance is $\mathrm{C}_{\mathrm{eq}}=\mathrm{C}(\mathrm{n}-1)$ $=3(91-1)$ $=270 \mathrm{pF}$
165796
An infinite number of capacitors each of capacity $2 \mu \mathrm{F}$ are connected as shown in the figure. The resultant capacity between $A$ and $B$ is
1 $5 \mu \mathrm{F}$
2 $1 \mu \mathrm{F}$
3 $3 \mu \mathrm{F}$
4 $4 \mu \mathrm{F}$
Explanation:
: Equivalent capacitance of Capacitors which are in series, $\frac{1}{\mathrm{C}_{\text {eq }}}=\frac{1}{\mathrm{C}_{1}}+\frac{1}{\mathrm{C}_{2}}+\ldots \ldots \ldots .+\frac{1}{\mathrm{C}_{\mathrm{n}}}$ For $\mathrm{C}_{1}=\mathrm{C}_{2}=\ldots \ldots \ldots \mathrm{C}_{\mathrm{n}}=2 \mu \mathrm{F}$ $\mathrm{C}_{\mathrm{eq}}=\frac{\mathrm{C}}{\mathrm{n}}$ For first row $\mathrm{C}_{\mathrm{A}}=\mathrm{C}$ For second row $\mathrm{C}_{\mathrm{B}}=\frac{\mathrm{C}}{2}$ For third row $\mathrm{C}_{\mathrm{C}}=\frac{\mathrm{C}}{4}$ and so on . From figure all the capacitor are in parallel. $\mathrm{C}_{\text {final }}=\mathrm{C}_{\mathrm{A}}+\mathrm{C}_{\mathrm{B}}+\mathrm{C}_{\mathrm{C}}+\ldots \ldots \ldots+\mathrm{C}_{\infty}$ $\mathrm{C}_{\text {final }}=\mathrm{C}+\frac{\mathrm{C}}{2}+\frac{\mathrm{C}}{4}+\frac{\mathrm{C}}{8}+\ldots \ldots \ldots+\frac{\mathrm{C}}{\infty}$ $\mathrm{C}_{\text {final }}=\frac{\mathrm{C}}{1-\frac{1}{2}}=2 \mathrm{C}$ $\mathrm{C}_{\text {final }}=2 \times 2 \times 10^{-6} \quad(\because \mathrm{C}=2 \mu \mathrm{F})$ $\mathrm{C}_{\text {final }}=4 \mu \mathrm{F}$
[AP EAMCET-28.04.2017
Capacitance
165797
Three capacitors each of capacitance $\mathbf{C}$ and of breakdown voltage $V$ are joined in series. The capacitance and breakdown voltage of the combination will be
1 $\frac{\mathrm{C}}{3}, \frac{\mathrm{V}}{3}$
2 $3 \mathrm{C}, \frac{\mathrm{V}}{3}$
3 $\frac{\mathrm{C}}{3}, 3 \mathrm{~V}$
4 $3 \mathrm{C}, 3 \mathrm{~V}$
Explanation:
: In series, combination of capacitors $\mathrm{V}_{\text {eff }}=\mathrm{V}+\mathrm{V}+\mathrm{V}=3 \mathrm{~V}$ And, $\frac{1}{\mathrm{C}_{\mathrm{eff}}}=\frac{1}{\mathrm{C}}+\frac{1}{\mathrm{C}}+\frac{1}{\mathrm{C}}$ $\frac{1}{C_{\text {eff }}}=\frac{3}{C}$ $C_{\text {eff }}=\frac{C}{3}$ Thus, the capacitance and break down voltage of the combination will be $\mathrm{C} / 3$ and $3 \mathrm{~V}$.
[AIPMT-2009]
Capacitance
165798
The equivalent capacitance of the combination shown in the figure is
1 $3 \mathrm{C}$
2 $2 \mathrm{C}$
3 $\frac{\mathrm{C}}{2}$
4 $\frac{3 \mathrm{C}}{2}$
Explanation:
: In this circuit one capacitor is shorted. So, the remaining two capacitor are in parallel. Given, $\mathrm{C}_{1}=\mathrm{C}_{2}=\mathrm{C}$ So, the equivalent capacitance is - $\mathrm{C}_{\mathrm{eq}}=\mathrm{C}_{1}+\mathrm{C}_{2}=2 \mathrm{C}$
[NEET-2021]
Capacitance
165799
A parallel plate capacitor has 91 plates, all are identical and arranged with same spacing between them. If the capacitance between adjacent plates is $3 \mathrm{pF}$. What will be the resultant capacitance?
1 $273 \mathrm{pF}$
2 $30 \mathrm{pF}$
3 $94 \mathrm{pF}$
4 $270 \mathrm{pF}$
Explanation:
: Given, $\mathrm{n}=91, \mathrm{C}=3 \mathrm{pF}$ When $\mathrm{n}$ plates are arranged with the same spacing between them, the combination behaves as $(n-1)$ capacitor in parallel since charge on each plate is same. So, equivalent capacitance is $\mathrm{C}_{\mathrm{eq}}=\mathrm{C}(\mathrm{n}-1)$ $=3(91-1)$ $=270 \mathrm{pF}$
