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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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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Effective Capacitance
Effective capacitance refers to the overall capacitance of a circuit that contains multiple capacitors. When capacitors are combined in a circuit, their combined capacitance behaves differently depending on whether they are connected in series or parallel. For circuits involving capacitors, it is crucial to understand how the effective capacitance is calculated.

For capacitors in series, the effective capacitance decreases compared to the capacitance of individual capacitors. This is because the inverses of the individual capacitances are summed up:
\[ \frac{1}{C_\mathrm{series}} = \frac{1}{C_1} + \frac{1}{C_2} \]
In this equation, \(C_\mathrm{series}\) is the effective capacitance, and \(C_1\) and \(C_2\) are the individual capacitances. This relationship means that the total capacitance in series is always less than the smallest individual capacitor in the series arrangement.

On the other hand, for capacitors in parallel, the effective capacitance is simply the sum of the individual capacitances:
\[ C_\mathrm{parallel} = C_1 + C_2 \]
This straightforward addition indicates that the total capacitance in parallel is greater than any single capacitor's capacitance, offering a larger storage for electrical charge. Understanding these basic principles is key to solving capacitance-related problems.
Series and Parallel Circuits
In electrical circuits, capacitors can be arranged in two fundamental ways: series and parallel. Each arrangement has unique characteristics that affect both the total capacitance and how the circuit functions.

  • Series Circuits: Capacitors in series are connected end-to-end, which means the charge on each capacitor is the same. However, the voltage across each is different. Because the total capacitance is found using the inverse sum rule, the effective capacitance is less than that of any single component. This arrangement is particularly useful when a lower capacitance is desired from higher-capacitance units.

  • Parallel Circuits: In a parallel arrangement, the capacitors connect across the same two points, so they each hold the same voltage but can have different charges. Adding capacitors in parallel increases the net capacitance since their individual capacitance values are added together. This is beneficial when additional electrical charge storage is needed.
Understanding these arrangements is critical for designing and analyzing circuits, as they determine how energy is stored and distributed across the circuit. Properly applying series and parallel configurations can significantly enhance a circuit's performance and efficiency.
Capacitance Formulas
To determine effective capacitance in various configurations, one must understand the basic formulas that govern capacitive circuits. Knowing when and how to apply these formulas is essential for accurately computing the total capacitance.

  • For capacitors connected in series, use the reciprocal formula:
    \[ \frac{1}{C_\mathrm{series}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} + \ldots \]
    This formula reflects how, in series, the overall capacitance decreases as more capacitors are added.

  • For capacitors in a parallel configuration, the total capacitance is the sum of individual capacitances:
    \[ C_\mathrm{parallel} = C_1 + C_2 + C_3 + \ldots \]
    In parallel, adding more capacitors increases the circuit's total capacitance, effectively increasing its energy storage capacity.
Mastery of these formulas allows for the analysis and design of a wide range of electrical circuits, allowing engineers and students to predict how a circuit will behave when different components are connected. Whether simplifying complex circuits or designing new ones, understanding these fundamental laws is vital.

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Most popular questions from this chapter

64 identical drops of mercury are charged simultaneously to the same potential of 10 volt. Assuming the drops to be spherical, if all the charged drops are made to combine to form one large drop, then its potential will be (A) \(100 \mathrm{~V}\) (B) \(320 \mathrm{~V}\) (C) \(640 \mathrm{~V}\) (D) \(160 \mathrm{~V}\)

A parallel plate capacitor has plate of area \(\mathrm{A}\) and separation d. It is charged to a potential difference \(\mathrm{V}_{0}\). The charging battery is disconnected and the plates are pulled apart to three times the initial separation. The work required to separate the plates is (A) \(\left[\left(\mathrm{A} \in_{0} \mathrm{~V}_{0}^{2}\right) / \mathrm{d}\right]\) (B) \(\left[\left(\mathrm{A} \in{ }_{0} \mathrm{~V}_{0}^{2}\right) /(2 \mathrm{~d})\right]\) (C) \(\left[\left(\mathrm{A} \in{ }_{0} \mathrm{~V}_{0}{ }^{2}\right) /(3 \mathrm{~d})\right]\) (D) \(\left[\left(\mathrm{A} \in{ }_{0} \mathrm{~V}_{0}{ }^{2}\right) /(4 \mathrm{~d})\right]\)

An oil drop of 12 excess electrons is held stationary under a constant electric field of \(2.55 \times 10^{4} \mathrm{Vm}^{-1}\). If the density of the oil is \(1.26 \mathrm{gm} / \mathrm{cm}^{3}\) then the radius of the drop is \(\ldots \ldots \ldots \mathrm{m}\). (A) \(9.75 \times 10^{-7}\) (B) \(9.29 \times 10^{-7}\) (C) \(9.38 \times 10^{-8}\) (D) \(9.34 \times 10^{-8}\)

Two identical balls having like charges and placed at a certain distance apart repel each other with a certain force. They are brought in contact and then moved apart to a distance equal to half their initial separation. The force of repulsion between them increases \(4.5\) times in comparison with the initial value. The ratio of the initial charges of the balls is ....... (A) \(4: 1\) (B) \(6: 1\) (C) \(3: 1\) (D) \(2: 1\)

The electric potential \(\mathrm{V}\) is given as a function of distance \(\mathrm{x}\) (meter) by \(\mathrm{V}=\left(5 \mathrm{x}^{2}+10 \mathrm{x}-9\right)\) volt. Value of electric field at \(\mathrm{x}=1\) is \(\ldots \ldots\) \((\mathrm{A})-20(\mathrm{v} / \mathrm{m})\) (B) \(6(\mathrm{v} / \mathrm{m})\) (C) \(11(\mathrm{v} / \mathrm{m})\) (D) \(-23(\mathrm{v} / \mathrm{m})\)

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