A $16\ \Omega$ wire is bend to form a square loop. A $9 \mathrm{~V}$ battery with internal resistance $1\ \Omega$ is connected across one of its sides. If a $4\ \mu \mathrm{F}$ capacitor is connected across one of its diagonals, the energy stored by the capacitor will be $\frac{x}{2} \ \mu \mathrm{J}$. where $x=$________.
JEE MAIN 2024, Diffcult
Download our app for free and get started
$ I=\frac{V}{R_{e q}} I=\frac{V}{R_{e q}}=\frac{9}{1+\frac{12 \times 4}{12+4}}=\frac{9}{4} $
$ I_1=\frac{9}{4} \times \frac{4}{16}=\frac{9}{16} $
$ V_A-V_B=I_1 \times 8=\frac{9}{16} \times 8=\frac{9}{2} V $
$ \therefore U=\frac{1}{2} \times 4 \times \frac{81}{4} \mu \mathrm{J} $
$ \therefore U=\frac{81}{2} \mu \mathrm{J} $
$ \therefore \mathrm{X}=81$
Download our appand get started for free
Experience the future of education. Simply download our apps or reach out to us for more information. Let's shape the future of learning together!No signup needed.*
Similar Questions
- 1The value of equivalent capacitance of the combination shown in figure between the points $P$ and $Q$ isView Solution
- 2Point charge ${q_1} = 2\,\mu C$ and ${q_2} = - 1\,\mu C$ are kept at points $x = 0$ and $x = 6$ respectively. Electrical potential will be zero at pointsView Solution
- 3Three capacitors each of $6\,\mu F$ are available. The minimum and maximum capacitances which may be obtained areView Solution
- 4A capacitor of $4\, \mu F$ is connected to a $15\, V$ supply through $1$ mega ohm resistance. The time taken by the capacitor to charge upto $63.2\%$ of its final charge will be......$s$View Solution
- 5Four electric charges $+q,+q, -q$ and $-q$ are placed at the comers of a square of side $2L$ (see figure). The electric potential at point $A,$ midway between the two charges $+q$ and $+q,$ isView Solution
- 6A charged capacitor is allowed to discharge through a resistor by closing the key at the instant $t =0$. At the instant $t = (ln \,4) $ $\mu s$, the reading of the ammeter falls half the initial value. The resistance of the ammeter is equal toView Solution
- 7A source of potential difference $V$ is connected to the combination of two identical capacitors as shown in the figure. When key ' $K$ ' is closed, the total energy stored across the combination is $E _{1}$. Now key ' $K$ ' is opened and dielectric of dielectric constant 5 is introduced between the plates of the capacitors. The total energy stored across the combination is now $E _{2}$. The ratio $E _{1} / E _{2}$ will be :View Solution
- 8In space of horizontal $EF$ ($E = (mg)/q$) exist as shown in figure and a mass $m$ attached at the end of a light rod. If mass $m$ is released from the position shown in figure find the angular velocity of the rod when it passes through the bottom most positionView Solution
- 9View SolutionAssertion: Electron move away from a region of higher potential to a region of lower potential.
Reason: An electron has a negative charge.
- 10A spherical drop of mercury having a potential of $2.5\, V$ is obtained as a result of merging $125$ droplets. The potential of constituent droplets would be........$V$View Solution