Question:
In the circuit shown, charge on the \(5\ \mu\mathrm{F}\) capacitor is :
In the circuit shown, charge on the \(5\ \mu\mathrm{F}\) capacitor is :
Show Hint
The 2 uF, 4 uF and 5 uF capacitors all join the same top and bottom nodes; each of the 2 uF and 4 uF is in series with a 6 V cell of the same orientation. Apply charge neutrality at the top node.
Updated On: Nov 5, 2026
- \(5.45\ \mu\mathrm{C}\)
- \(18.00\ \mu\mathrm{C}\)
- \(10.90\ \mu\mathrm{C}\)
- \(16.36\ \mu\mathrm{C}\)
Show Solution
Verified By Collegedunia
The Correct Option is D
Solution and Explanation
Let the top-middle node be X and the bottom-middle node be Y. Both 6 V cells have their negative plate towards Y (short plate at the middle) and their positive plate on the outer side, so they act in the same sense. The 2 \(\mu\)F, 4 \(\mu\)F and 5 \(\mu\)F capacitors all connect X to Y, with 6 V cells in series with the first two.
Let \(V = V_X - V_Y\). Net charge on the isolated node X is zero:
\(2(V - 6) + 4(V - 6) + 5V = 0\)
\(11V = 36 \Rightarrow V = \frac{36}{11}\ \mathrm{V} = 3.27\ \mathrm{V}\).
Charge on the 5 \(\mu\)F capacitor: \(Q = 5 \times \frac{36}{11} = 16.36\ \mu\mathrm{C}\).
Answer: (4) 16.36 \(\mu\)C
Let \(V = V_X - V_Y\). Net charge on the isolated node X is zero:
\(2(V - 6) + 4(V - 6) + 5V = 0\)
\(11V = 36 \Rightarrow V = \frac{36}{11}\ \mathrm{V} = 3.27\ \mathrm{V}\).
Charge on the 5 \(\mu\)F capacitor: \(Q = 5 \times \frac{36}{11} = 16.36\ \mu\mathrm{C}\).
Answer: (4) 16.36 \(\mu\)C
Was this answer helpful?
0
0
Top JEE Main Physics Questions
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,
- $I _{ CM }$ is the moment of inertia of a circular disc about an axis (CM) passing through its center and perpendicular to the plane of disc $I_{A B}$ is it's moment of inertia about an axis $AB$ perpendicular to plane and parallel to axis $CM$ at a distance $\frac{2}{3} R$ from center Where $R$ is the radius of the dise The ratio of $I _{ AB }$ and $I _{ CM }$ is $x: 9$ The value of $x$ is______
- A nucleus disintegrates into two smaller parts, which have their velocities in the ratio $3: 2$ The ratio of their nuclear sizes will be $\left(\frac{x}{3}\right)^{\frac{1}{3}}$ The value of ' $x$ ' is:-
- A particle starts with an initial velocity of 10.0 m/s along x-direction and accelerates uniformly at the rate of 2.0 m/s$^2$. The time taken by the particle to reach the velocity of 60.0 m/s is
View More Questions
Top JEE Main Electrostatics Questions
- The resistivity (\( \rho \)) of a semiconductor varies with temperature. Which of the following curves represents the correct behavior?
- Given below are two statements : one is labelled a Assertion (A) and the other is labelled as Reason(R)
Assertion (A) : Work done by electric field on moving a positive charge on an equipotential surface is always zero.
Reason (R) : Electric lines of forces are always perpendicular to equipotential surfaces.
In the light of the above statements, choose the most appropriate answer from the options given below : - Two charges of \( -4 \, \mu C \) and \( +4 \, \mu C \) are placed at the points \( A(1, 0, 4) \, m \) and \( B(2, -1, 5) \, m \) located in an electric field \( \vec{E} = 0.20 \, \hat{i} \, V / cm \). The magnitude of the torque acting on the dipole is \( 8\sqrt{\alpha} \times 10^{-5} \, Nm \), where \( \alpha = \, \).
- An electron is moving under the influence of the electric field of a uniformly charged infinite plane sheet \( S \) having surface charge density \( +\sigma \). The electron at \( t = 0 \) is at a distance of 1 m from \( S \) and has a speed of 1 m/s. The maximum value of \( \sigma \) if the electron strikes \( S \) at \( t = 1 \, \text{s} \) is \( \alpha \left[ \frac{m \epsilon_0}{e} \right] \frac{\text{C}}{\text{m}^2} \). The value of \( \alpha \) is ______.
- Two identical charged spheres are suspended by strings of equal lengths. The strings make an angle of 37° with each other. When suspended in a liquid of density 0.7 g/cm3, the angle remains the same. If the density of the material of the sphere is 1.4 g/cm3, the dielectric constant of the liquid is ___________. (tan 37° =\( \frac{3}{4}\))
View More Questions
Top JEE Main Questions
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,
View More Questions