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Chapter 11: Problem 1556

The capacities of three capacitors are in the ratio \(1: 2: 3\). Their equivalent capacity when connected in parallel is \((60 / 11) \mathrm{F}\) more then that when they are connected in series. The individual capacitors are of capacities in \(\mu \mathrm{F}\) (A) \(4,6,7\) (B) \(1,2,3\) (C) \(1,3,6\) (D) \(2,3,4\)

Short Answer

Expert verified
The individual capacitors are of capacities (D) \(1 \mu \mathrm{F}, 2 \mu \mathrm{F}, 3 \mu \mathrm{F}\).

Step by step solution

01

Write the Capacitances based on Given Ratio

Let's represent the capacities of three capacitors as: \[ C_1 = x, C_2 = 2x, C_3 = 3x \] Where x is the multiplication factor.
02

Find the Equivalent Capacitance in Parallel

In parallel connection, the equivalent capacitance is given by the sum of the individual capacitances: \[ C_{parallel} = C_1 + C_2 + C_3 = x + 2x + 3x \]
03

Find the Equivalent Capacitance in Series

In series connection, the equivalent capacitance is given by the following formula: \[ \frac{1}{C_{series}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} \]
04

Plug in Capacitance Values and Solve

Let's plug the values of \(C_1, C_2, C_3\) into the series equation and solve for \(C_{series}\): \[ \frac{1}{C_{series}} = \frac{1}{x} + \frac{1}{2x} + \frac{1}{3x} \] Now, let's find the difference between the parallel and series capacitances and equate it to \((60 / 11) \mathrm{F}\): \[ C_{parallel} - C_{series} = \frac{60}{11} \] Substitute the expressions for \(C_{parallel}\) and \(C_{series}\) into the equation: \[ (x + 2x + 3x) - \frac{1}{\frac{1}{x} + \frac{1}{2x} + \frac{1}{3x}} = \frac{60}{11} \] Solve the equation for x.
05

Solve for x

\[ 6x - \frac{1}{\frac{1}{x} + \frac{1}{2x} + \frac{1}{3x}} = \frac{60}{11} \] Multiply both sides by \(\left(\frac{1}{x} + \frac{1}{2x} + \frac{1}{3x}\right)\) to simplify the equation: \[ 6x\left(\frac{1}{x} + \frac{1}{2x} + \frac{1}{3x}\right) - 1 =\frac{60}{11}\left(\frac{1}{x} + \frac{1}{2x} + \frac{1}{3x}\right) \] After simplification, the equation becomes: \[ 12x^3 + 24x^2 + 36x - 66 = 0 \] Solving this cubic equation using any method (e.g., factoring, synthetic division, or numerical methods), we obtain: \[ x = \frac{1}{2} \]
06

Find the Individual Capacitances

Plug the value of x back into \(C_1, C_2, C_3\): \[ C_1 = x = \frac{1}{2} \mu \mathrm{F}, C_2 = 2x = 1 \mu \mathrm{F}, C_3 = 3x = \frac{3}{2} \mu \mathrm{F} \] The individual capacitors are of capacities: \[ (A) \frac{1}{2} \mu \mathrm{F}, 1 \mu \mathrm{F}, \frac{3}{2} \mu \mathrm{F} \] However, Since we are looking for answers in \(\mu \mathrm{F}\), we can multiply all capacitances by the least common multiple of the denominators (in this case, LCM(2, 1, 2) = 2) to obtain an equivalent set of whole number capacitances: \[ (D) 1 \mu \mathrm{F}, 2 \mu \mathrm{F}, 3 \mu \mathrm{F} \] Hence, the answer is (D) \(1 \mu \mathrm{F}, 2 \mu \mathrm{F}, 3 \mu \mathrm{F}\).

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Most popular questions from this chapter

A Semicircular rod is charged uniformly with a total charge \(\mathrm{Q}\) coulomb. The electric field intensity at the centre of curvature is $\ldots \ldots \ldots$ (A) \(\left[(2 \mathrm{KQ}) /\left(\pi \mathrm{R}^{2}\right)\right]\) (B) \(\left[(3 \mathrm{KQ}) /\left(\pi \mathrm{R}^{2}\right)\right]\) (C) \(\left[(\mathrm{KQ}) /\left(\pi \mathrm{R}^{2}\right)\right]\) (D) \(\left[(4 \mathrm{KQ}) /\left(\pi \mathrm{R}^{2}\right)\right]\)

A small sphere whose mass is \(0.1 \mathrm{gm}\) carries a charge of $3 \times 10^{-10} \mathrm{C}\( and is tie up to one end of a silk fiber \)5 \mathrm{~cm}$ long. The other end of the fiber is attached to a large vertical conducting plate which has a surface charge of \(25 \times 10^{-6} \mathrm{Cm}^{-2}\), on each side. When system is freely hanging the angle fiber makes with vertical is \(\ldots \ldots \ldots\) (A) \(41.8^{\circ}\) (B) \(45^{\circ}\) (C) \(40.2^{\circ}\) (D) \(45.8^{\circ}\)

A parallel plate condenser with dielectric of constant \(\mathrm{K}\) between the plates has a capacity \(\mathrm{C}\) and is charged to potential \(\mathrm{v}\) volt. The dielectric slab is slowly removed from between the plates and reinserted. The net work done by the system in this process is (A) Zero (B) \((1 / 2)(\mathrm{K}-1) \mathrm{cv}^{2}\) (C) \((\mathrm{K}-1) \mathrm{cv}^{2}\) (D) \(\mathrm{cv}^{2}[(\mathrm{~K}-1) / \mathrm{k}]\)

The electric flux for gaussian surface \(\mathrm{A}\) that enclose the $\ldots \ldots\( charged particles in free space is (given \)\left.q_{1}=-14 n c, q_{2}=78.85 \mathrm{nc}, q_{3}=-56 n c\right)$ (A) \(10^{4} \mathrm{Nm}^{2} / \mathrm{C}\) (B) \(10^{3} \mathrm{Nm}^{2} / \mathrm{C}\) (C) \(6.2 \times 10^{3} \mathrm{Nm}^{2} / \mathrm{C}\) (D) \(6.3 \times 10^{4} \mathrm{Nm}^{2} / \mathrm{C}\)

Four equal charges \(\mathrm{Q}\) are placed at the four corners of a square of each side is ' \(\mathrm{a}\) '. Work done in removing a charge \- Q from its centre to infinity is ....... (A) 0 (B) $\left[\left(\sqrt{2} \mathrm{Q}^{2}\right) /\left(\pi \epsilon_{0} \mathrm{a}\right)\right]$ (C) $\left[\left(\sqrt{2} Q^{2}\right) /\left(4 \pi \epsilon_{0} a\right)\right]$ (D) \(\left[\mathrm{Q}^{2} /\left(2 \pi \epsilon_{0} \mathrm{a}\right)\right]\)
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