Physics-

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### Question

#### Effective capacitance between points in the figure, shown is

#### The correct answer is:

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### Related Questions to study

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#### A number of capacitors, each of equal capacitance , are arranged as shown in Fig. The equivalent capacitance between and is

#### A number of capacitors, each of equal capacitance , are arranged as shown in Fig. The equivalent capacitance between and is

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#### Two capacitors $A$ and $B$ with capacities ${C}_{1}$ and ${C}_{2}$ are charged to potential difference of ${V}_{1}$and${V}_{2}$, respectively. The plates of the capacitors are connected as shown in the figure with one wire free from each capacitor. The upper plate of $A$ is positive and that of $B$ is negative. An uncharged capacitor of capacitance ${C}_{3}$ and lead wires falls on the free ends to complete circuit, then

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#### Find the charge that will flow through the wire $A$ to $B$ if the switch $S$ is closed. The capacitance of each capacitor shown in the figure is$C$

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#### A 20$\mu F$ capacitor is connected to 45 V battery through a circuit whose resistance is 2000$\Omega .$ What is the final charge on the capacitor?

We know that in steady state the capacitor behaves like as an open circuit $ie$, capacitor will not pass the current.
So, the potential difference across the capacitor = 45 V
Hence , the final charge on the capacitor is
$q=CV$
Here , $C=20\mu F,V=45V$
$\therefore q=20\times {10}^{-6}\times 45$
or$q=900\times {10}^{-6}$
or$q=9\times {10}^{-4}C$

#### A 20$\mu F$ capacitor is connected to 45 V battery through a circuit whose resistance is 2000$\Omega .$ What is the final charge on the capacitor?

physics-General

We know that in steady state the capacitor behaves like as an open circuit $ie$, capacitor will not pass the current.
So, the potential difference across the capacitor = 45 V
Hence , the final charge on the capacitor is
$q=CV$
Here , $C=20\mu F,V=45V$
$\therefore q=20\times {10}^{-6}\times 45$
or$q=900\times {10}^{-6}$
or$q=9\times {10}^{-4}C$

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#### Assertion (A):If $a>0,b>0$ and $c>0$, then $(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\ge 9$Reason (R): For positive numbers $a,$ $b,$ $c,$ $AM\ge GM$

$AM\ge GM$$=\succ \frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}{3}\ge \left(\begin{array}{cc}111& \\ -& -\\ \overline{b}a.& c\end{array}\right)\to \left(1\right)$$D:{x}^{2}-3x-4<0\supset (x+1)(x-4)<0\Rightarrow 1<x<4$${x}^{2}-3x+2>0Rejectx-2\left)\right(x-1)>0=x<1{a}^{r}x>2$$\frac{(a+b+c)}{3}\ge (abc{)}^{\frac{1}{3}}\to (2)$multiply (1) and (2)

#### Assertion (A):If $a>0,b>0$ and $c>0$, then $(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\ge 9$Reason (R): For positive numbers $a,$ $b,$ $c,$ $AM\ge GM$

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$AM\ge GM$$=\succ \frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}{3}\ge \left(\begin{array}{cc}111& \\ -& -\\ \overline{b}a.& c\end{array}\right)\to \left(1\right)$$D:{x}^{2}-3x-4<0\supset (x+1)(x-4)<0\Rightarrow 1<x<4$${x}^{2}-3x+2>0Rejectx-2\left)\right(x-1)>0=x<1{a}^{r}x>2$$\frac{(a+b+c)}{3}\ge (abc{)}^{\frac{1}{3}}\to (2)$multiply (1) and (2)

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#### The length of the chord joining the points $\left(4\mathrm{cos}\theta ,4\mathrm{sin}\theta \right)$ and $(4\mathrm{cos}\left(\theta +60\xb0\right),4\mathrm{sin}(\theta +60\xb0)$of the circle ${x}^{2}+{y}^{2}=16$ is

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