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Chapter 11: Problem 1611

The eftective capacitances of two capacitors are \(3 \mu \mathrm{F}\) and $16 \mu \mathrm{F}$, when they are connected in series and parallel respectively. The capacitance of each capacitor is (A) \(2 \mu \mathrm{F}, 14 \mu \mathrm{F}\) (B) \(4 \mu \mathrm{F}, 12 \mu \mathrm{F}\) (C) \(6 \mu \mathrm{F}, 8 \mu \mathrm{F}\) (D) \(10 \mu \mathrm{F}, 6 \mu \mathrm{F}\)

Short Answer

Expert verified
The capacitance of each capacitor is (B) \(4 \mu\mathrm{F}\), \(12 \mu\mathrm{F}\).

Step by step solution

01

Write down the general formula for effective capacitance in series

We'll first write down the general formula for effective capacitance in series, which is: \[ \frac{1}{C_\mathrm{series}} = \frac{1}{C_1} + \frac{1}{C_2} \] where \(C_\mathrm{series}\) is the effective capacitance in series, \(C_1\) is the capacitance of the first capacitor, and \(C_2\) is the capacitance of the second capacitor.
02

Write down the general formula for effective capacitance in parallel

Now, let's write the general formula for effective capacitance in parallel, which is: \[ C_\mathrm{parallel} = C_1 + C_2 \] where \(C_\mathrm{parallel}\) is the effective capacitance in parallel.
03

Plug in the given effective capacitance values

From the exercise, we're given that \(C_\mathrm{series} = 3 \mu\mathrm{F}\), and \(C_\mathrm{parallel} = 16 \mu\mathrm{F}\). Plugging these values into the formulas from Steps 1 and 2, we get: \[ \frac{1}{3\mu\mathrm{F}} = \frac{1}{C_1} + \frac{1}{C_2} \] \[ 16\mu\mathrm{F} = C_1 + C_2 \]
04

Solve for the capacitance values

Now, we have two equations with two unknowns, \(C_1\) and \(C_2\). We will solve these equations simultaneously to find the values of \(C_1\) and \(C_2\). One method to solve these equations is by substitution. From the parallel equation, we can express \(C_1\) as: \[ C_1 = 16\mu\mathrm{F} - C_2 \] Now substitute this expression for \(C_1\) into the series equation: \[ \frac{1}{3\mu\mathrm{F}} = \frac{1}{16\mu\mathrm{F} - C_2} + \frac{1}{C_2} \] Rearrange the terms: \[ 3\mu\mathrm{F}C_2(16\mu\mathrm{F} - C_2) = (16\mu\mathrm{F} - C_2) C_2 \] Now, we can plug in the given options for \(C_1\) and \(C_2\) into this equation to see which option satisfies the equation.
05

Test the given options

The given options are: (A) \(2 \mu\mathrm{F}\), \(14 \mu\mathrm{F}\) (B) \(4 \mu\mathrm{F}\), \(12 \mu\mathrm{F}\) (C) \(6 \mu\mathrm{F}\), \(8 \mu\mathrm{F}\) (D) \(10 \mu\mathrm{F}\), \(6 \mu\mathrm{F}\) The only option that satisfies the equation is: \[ 3\mu\mathrm{F} \cdot 4\mu\mathrm{F} \cdot (16\mu\mathrm{F} - 4\mu\mathrm{F}) = (16\mu\mathrm{F} - 4\mu\mathrm{F}) \cdot 4\mu\mathrm{F} \] So, the correct answer is (B) \(4 \mu\mathrm{F}\), \(12 \mu\mathrm{F}\).

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Most popular questions from this chapter

A parallel plate capacitor of capacitance \(5 \mu \mathrm{F}\) and plate separation \(6 \mathrm{~cm}\) is connected to a \(1 \mathrm{~V}\) battery and charged. A dielectric of dielectric constant 4 and thickness \(4 \mathrm{~cm}\) is introduced between the plates of the capacitor. The additional charge that flows into the capacitor from the battery is (A) \(2 \mu \mathrm{c}\) (B) \(5 \mu \mathrm{c}\) (C) \(3 \mu \mathrm{c}\) (D) \(10 \mu \mathrm{c}\)

Charges \(+q\) and \(-q\) are placed at point \(A\) and \(B\) respectively which are a distance \(2 \mathrm{~L}\) apart, \(\mathrm{C}\) is the midpoint between \(\mathrm{A}\) and \(\mathrm{B}\). The work done in moving a charge \(+Q\) along the semicircle \(C R D\) is \(\ldots \ldots\) (A) $\left[(\mathrm{qQ}) /\left(2 \pi \mathrm{\epsilon}_{0} \mathrm{~L}\right)\right]$ (B) \(\left[(-q Q) /\left(6 \pi \in_{0} L\right)\right]\) (C) $\left[(\mathrm{qQ}) /\left(6 \pi \mathrm{e}_{0} \mathrm{~L}\right)\right]$ (D) \(\left[(q Q) /\left(4 \pi \in_{0} L\right)\right]\)

A small conducting sphere of radius \(r\) is lying concentrically inside a bigger hollow conducting sphere of radius \(R\). The bigger and smaller sphere are charged with \(\mathrm{Q}\) and \(\mathrm{q}(\mathrm{Q}>\mathrm{q})\) and are insulated from each other. The potential difference between the spheres will be \(\ldots \ldots\) (A) \(\left[1 /\left(4 \pi \epsilon_{0}\right)\right][(q / r)-(Q / R)]\) (B) \(\left[1 /\left(4 \pi \epsilon_{0}\right)\right][(q / r)-(q / R)]\) (C) \(\left[1 /\left(4 \pi \epsilon_{0}\right)\right][(Q / R)+(q / r)]\) (D) $\left[1 /\left(4 \pi \epsilon_{0}\right)\right][(\mathrm{q} / \mathrm{R})-(\mathrm{Q} / \mathrm{r})]$

Iwo points are at distances a and b \((a

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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