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Chapter 11: Problem 1622

The capacitors of capacitance \(4 \mu \mathrm{F}, 6 \mu \mathrm{F}\) and $12 \mu \mathrm{F}$ are connected first in series and then in parallel. What is the ratio of equivalent capacitance in the two cases? (A) \(2: 3\) (B) \(11: 1\) (C) \(1: 11\) (D) \(1: 3\)

Short Answer

Expert verified
The ratio of equivalent capacitance in the two cases is \(1: 11\) (Option C).

Step by step solution

01

To find the equivalent capacitance (\(C_s\)) in a series configuration, we use the following formula: \[\frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}\] where \(C_1 = 4 \mu F\), \(C_2 = 6 \mu F\), and \(C_3 = 12 \mu F\). #Step 2: Substitute values and solve for equivalent capacitance in series#

Substitute the values of the capacitors and solve for \(C_s\): \[\frac{1}{C_s} = \frac{1}{4} + \frac{1}{6} + \frac{1}{12}\] #Step 3: Simplify the equation for Cs#
02

To simplify the equation, we can find the least common denominator of \(4\), \(6\), and \(12\), as follows: \[\frac{1}{C_s} = \frac{3+2+1}{12}\] #Step 4: Solve for Cs#

Solve for \(C_s\): \[\frac{1}{C_s} = \frac{6}{12}\] \[C_s = 2 \mu F\] #Step 5: Calculate the equivalent capacitance in parallel configuration#
03

To find the equivalent capacitance (\(C_p\)) in a parallel configuration, we use the following formula: \[C_p = C_1 + C_2 + C_3\] #Step 6: Substitute values and solve for equivalent capacitance in parallel#

Substitute the values of the capacitors and solve for \(C_p\): \[C_p = 4 + 6 + 12\] \[C_p = 22 \mu F\] #Step 7: Calculate the ratio of equivalent capacitance in series and parallel configurations#
04

Now that we have the equivalent capacitance in series and parallel configurations, we can find the ratio between these values: \[\text{Ratio}= \frac{C_s}{C_p}\] #Step 8: Substitute values and solve for ratio#

Substitute the values of \(C_s\) and \(C_p\) and solve for the ratio: \[\text{Ratio}= \frac{2}{22}\] \[\text{Ratio}= \frac{1}{11}\] Therefore, the ratio of equivalent capacitance in the two cases is \(\boldsymbol{1: 11}\) (Option C).

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Most popular questions from this chapter

Three particles, each having a charge of \(10 \mu \mathrm{c}\) are placed at the corners of an equilateral triangle of side \(10 \mathrm{~cm}\). The electrostatic potential energy of the system is (Given $\left.\left[1 /\left(4 \pi \epsilon_{0}\right)\right]=9 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{c}^{2}\right)$ (A) \(100 \mathrm{~J}\) (B) \(27 \mathrm{~J}\) (C) Zero (D) Infinite

For the system shown in figure, if the resultant force on q is zero, then \(q=\ldots \ldots \ldots\) (A) \(-2 \sqrt{2} \mathrm{Q}\) (B) \(2 \sqrt{2} \mathrm{Q}\) (C) \(2 \sqrt{3} \mathrm{Q}\) (D) \(-3 \sqrt{2} Q\)

At what angle \(\theta\) a point \(P\) must be located from dipole axis so that the electric field intensity at the point is perpendicular to the dipole axis? (A) \(\tan ^{-1}(1 / \sqrt{2})\) (B) \(\tan ^{-1}(1 / 2)\) (C) \(\tan ^{-1}(2)\) (C) \(\tan ^{-1}(\sqrt{2})\)

Two air capacitors \(A=1 \mu F, B=4 \mu F\) are connected in series with $35 \mathrm{~V}\( source. When a medium of dielectric constant \)\mathrm{K}=3$ is introduced between the plates of \(\mathrm{A}\), change on the capacitor changes by (A) \(16 \mu \mathrm{c}\) (B) \(32 \mu \mathrm{c}\) (C) \(28 \mu \mathrm{c}\) (D) \(60 \mu \mathrm{c}\)

Two thin wire rings each having a radius \(R\) are placed at a distance \(d\) apart with their axes coinciding. The charges on the two rings are \(+q\) and \(-q\). The potential difference between the centers of the two rings is $\ldots .$ (A) 0 (B) $\left.\left[\mathrm{q} /\left(2 \pi \epsilon_{0}\right)\right]\left[(1 / \mathrm{R})-\left\\{1 / \sqrt{(}^{2}+\mathrm{d}^{2}\right)\right\\}\right]$ (C) $\left[\mathrm{q} /\left(4 \pi \epsilon_{0}\right)\right]\left[(1 / \mathrm{R})-\left\\{1 / \sqrt{\left. \left.\left(\mathrm{R}^{2}+\mathrm{d}^{2}\right)\right\\}\right]}\right.\right.$ (D) \(\left[(q R) /\left(4 \pi \epsilon_{0} d^{2}\right)\right]\)
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