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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

Capacitance of a parallel plate capacitor becomes \((4 / 3)\) times its original value if a dielectric slab of thickness \(t=d / 2\) is inserted between the plates (d is the separation between the plates). The dielectric constant of the slab is (A) 8 (B) 4 (C) 6 (D) 2

Two point charges \(-q\) and \(+q\) are located at points \((0,0,-a)\) and $(0,0, a)\( respectively. The potential at a point \)(0,0, z)\( where \)z>a\( is \)\ldots \ldots$ (A) $\left[(2 \mathrm{q} a) /\left\\{4 \pi \epsilon_{0}\left(z^{2}+a^{2}\right)\right\\}\right]$ (B) \(\left[\mathrm{q} /\left(4 \pi \epsilon_{0} \mathrm{a}\right)\right]\) (C) \(\left[\right.\) (qa) \(\left./\left(4 \pi \in_{0} z^{2}\right)\right]\) (D) $\left[(2 q a) /\left\\{4 \pi \epsilon_{0}\left(z^{2}-a^{2}\right)\right\\}\right]$

A \(5 \mu \mathrm{F}\) capacitor is charged by a \(220 \mathrm{v}\) supply. It is then disconnected from the supply and is connected to another uncharged $2.5 \mu \mathrm{F}$ capacitor. How much electrostatic energy of the first capacitor is lost in the form of heat and electromagnetic radiation ? (A) \(0.02 \mathrm{~J}\) (B) \(0.121 \mathrm{~J}\) (C) \(0.04 \mathrm{~J}\) (D) \(0.081 \mathrm{~J}\)

Two Points \(P\) and \(Q\) are maintained at the Potentials of \(10 \mathrm{v}\) and \(-4 \mathrm{v}\), respectively. The work done in moving 100 electrons from \(\mathrm{P}\) to \(\mathrm{Q}\) is \(\ldots \ldots \ldots\) (A) \(2.24 \times 10^{-16} \mathrm{~J}\) (B) \(-9.60 \times 10^{-17} \mathrm{~J}\) (C) \(-2.24 \times 10^{-16} \mathrm{~J}\) (D) \(9.60 \times 10^{-17} \mathrm{~J}\)

Two equal charges \(q\) are placed at a distance of \(2 \mathrm{a}\) and a third charge \(-2 q\) is placed at the midpoint. The potential energy of the system is ....... (A) $\left[\left(9 \mathrm{q}^{2}\right) /\left(8 \pi \epsilon_{0} \mathrm{a}\right)\right]$ (B) \(\left[q^{2} /\left(8 \pi \epsilon_{0} a\right)\right]\) (C) \(-\left[\left(7 q^{2}\right) /\left(8 \pi \epsilon_{0} a\right)\right]\) (D) \(\left[\left(6 q^{2}\right) /\left(8 \pi \in_{0} a\right)\right]\)
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