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Chapter 11: Problem 1658

An electric circuit requires a total capacitance of \(2 \mu \mathrm{F}\) across a potential of \(1000 \mathrm{~V}\). Large number of \(1 \mu \mathrm{F}\) capacitances are available each of which would breakdown if the potential is more then \(350 \mathrm{~V}\). How many capacitances are required to make the circuit? (A) 24 (B) 12 (C) 20 (D) 18

Short Answer

Expert verified
The total number of capacitors needed to make the circuit with a total capacitance of \(2\mu F\) and a potential of \(1000V\) using \(1\mu F\) capacitors with a maximum potential of \(350V\) is \(18\). The correct answer is (D) 18.

Step by step solution

01

Basic understanding of Capacitors in Parallel and Series

When capacitors are connected in parallel, their capacitances add up for a total capacitance, and the potential across each capacitor is the same. The formula for the total capacitance in parallel is: \[ C_\text{total} = C_1 + C_2 + \cdots + C_n \] When capacitors are connected in series, the total capacitance decreases, and the potential (voltage) across each capacitor adds up to the potential across the entire arrangement. The formula for the total capacitance in series is: \[ \frac{1}{C_\text{total}} = \frac{1}{C_1} + \frac{1}{C_2} + \cdots + \frac{1}{C_n} \]
02

Calculate the number of capacitors in series

Since each capacitor can handle a maximum potential of \(350V\), we need to determine the number of capacitors in series that can handle (at least) the \(1000V\) potential. Let \(n_\text{series}\) be the number of capacitors in series, and we have: \[ 350 \cdot n_\text{series} \geq 1000 \] Solve the inequality to find the minimum number of capacitors in series: \[ n_\text{series} \geq \frac{1000}{350} \approx 2.86 \] Since we can't have a non-integer number of capacitors, let's use the smallest integer greater or equal to \(2.86\), that is \(n_\text{series} = 3\). So, we need at least \(3\) capacitors connected in series to handle the \(1000V\) potential.
03

Calculate the capacitance of the series arrangement

Now that we have \(3\) capacitors in series, let's calculate the total capacitance of this arrangement. Each capacitor has a capacitance of \(1\mu F\). Using the series capacitance formula from step 1, we get: \[ \frac{1}{C_\text{series}} = \frac{1}{1} + \frac{1}{1} + \frac{1}{1} = 3 \] Solving this equation for the total capacitance of the series arrangement, we get: \[ C_\text{series} = \frac{1}{3}\mu F \]
04

Calculate the required number of capacitors in parallel

Our goal is to create a total capacitance of \(2\mu F\). We already found out that connecting the \(3\) capacitors in series gives us a total capacitance of \(\dfrac{1}{3}\mu F\). Let's call \(n_\text{parallel}\) the number of series arrangements needed in parallel to achieve the desired capacitance. Using the parallel capacitance equation from step 1, we get: \[ C_\text{total} = n_\text{parallel} \cdot C_\text{series} \implies 2 = n_\text{parallel} \cdot \frac{1}{3} \] Solving for the required number of capacitors in parallel: \[ n_\text{parallel} = \frac{2}{\frac{1}{3}} = 6 \]
05

Calculate the total number of capacitors used

Now that we have determined the number of capacitors needed in series and parallel to reach the desired total capacitance and handle the potential, we can calculate the total number of capacitors used in the arrangement. Multiplying the capacitors in series and in parallel: \[ n_\text{total} = n_\text{series} \cdot n_\text{parallel} = 3 \cdot 6 = 18 \] Therefore, the total number of capacitors needed is \(18\). The correct answer is (D) 18.

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Most popular questions from this chapter

A point charge \(q\) is situated at a distance \(r\) from one end of a thin conducting rod of length \(\mathrm{L}\) having a charge \(\mathrm{Q}\) (uniformly distributed along its length). The magnitude of electric force between the two, is ...... \((\mathrm{A})[(2 \mathrm{kqQ}) / \mathrm{r}(\mathrm{r}+\mathrm{L})]\) (B) \([(\mathrm{kq} \mathrm{Q}) / \mathrm{r}(\mathrm{r}+\mathrm{L})]\) (C) \([(\mathrm{kqQ}) / \mathrm{r}(\mathrm{r}-\mathrm{L})]\) (D) \([(\mathrm{kQ}) / \mathrm{r}(\mathrm{r}+\mathrm{L})]\)

A parallel plate air capacitor has a capacitance \(\mathrm{C}\). When it is half filled with a dielectric of dielectric constant 5, the percentage increase in the capacitance will be (A) \(200 \%\) (B) \(33.3 \%\) (C) \(400 \%\) (D) \(66.6 \%\)

Three charges, each of value \(Q\), are placed at the vertex of an equilateral triangle. A fourth charge \(q\) is placed at the centre of the triangle. If the charges remains stationery then, \(q=\ldots \ldots \ldots\) (A) \((\mathrm{Q} / \sqrt{2})\) (B) \(-(\mathrm{Q} / \sqrt{3})\) (C) \(-(Q / \sqrt{2})\) (D) \((\mathrm{Q} / \sqrt{3})\)

Point charges \(4 \mu \mathrm{c}\) and \(2 \mu \mathrm{c}\) are placed at the vertices \(\mathrm{P}\) and Q of a right angle triangle \(P Q R\) respectively. \(Q\) is the right angle, \(\mathrm{PR}=2 \times 10^{-2} \mathrm{~m}\) and \(\mathrm{QR}=10^{-2} \mathrm{~m}\). The magnitude and direction of the resultant electric field at \(\mathrm{R}\) is \(\ldots \ldots\) (A) \(4.28 \times 10^{9} \mathrm{NC}^{-1}, 45^{\circ}\) (B) \(2.38 \times 10^{8} \mathrm{NC}^{-1}, 40.9^{\circ}\) (C) \(1.73 \times 10^{4} \mathrm{NC}^{-1}, 34.7^{\circ}\) (D) \(4.9 \times 10^{10} \mathrm{NC}^{-1}, 34.7^{\circ}\)

Three charges \(2 q,-q,-q\) are located at the vertices of an equilateral triangle. At the centre of the triangle. (A) The Field is Zero but Potential is non - zero (B) The Field is non - Zero but Potential is zero (C) Both field and Potential are Zero (D) Both field and Potential are non - Zero
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