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The resistance of the series combination of two resistances is \(\mathrm{S}\), when they are joined in parallel the total resistance is \(\mathrm{P}\) If \(S=n P\), then the minimum possible value of \(n\) is.... (A) 4 (B) 3 (C) 2 (D) 1

Short Answer

Expert verified
The minimum possible value of \(n\) is 4. (A)

Step by step solution

01

Recall the formulas for calculating resistance in series and parallel

Let's denote the two resistances as \(R_1\) and \(R_2\). The formula for calculating the total resistance in series (\(S\)) and parallel (\(P\)) are as follows: For series: \(S = R_1 + R_2\) For parallel: \(\displaystyle P = \frac{R_{1} R_{2}}{R_{1} +R_{2}}\) We are given that \(S = nP\) and our task is to find the minimum value of \(n\).
02

Substitute the formulas for series and parallel into the equation

Replace the values of \(S\) and \(P\) with their respective formulas from Step 1 into the equation \(S = nP\): \(R_1 + R_2 = n \cdot \frac {R_1R_2}{R_1+R_2}\)
03

Simplify the equation and eliminate common terms

The equation from Step 2 can be simplified by multiplying both sides by \((R_1+R_2)\): \((R_1 + R_2)^2 = n R_1 R_2\) Now, expand the left side of the equation: \(R_1^2 + 2R_1R_2 + R_2^2 = n R_1 R_2\)
04

Minimize the value of n

To find the minimum value of \(n\), we will look for the minimum possible non-negative difference between the left side and right side of the equation. This occurs when the equation becomes an equality, which means the right term of the left side (\(R_1^2 + R_2^2\)) should be as small as possible. The smallest value for the sum of squares of two positive values (as resistances are always positive) occurs when both \(R_1\) and \(R_2\) are equal. In that case, \(R_1 = R_2\) Now, substitute this condition into the equation from Step 3: \((R_1 + R_1)^2 = n R_1 R_1\) \(4R_1^2 = n R_1^2\)
05

Solve for n

Now divide both sides by \(R_1^2\) to find the minimum possible value of n: \(n = \frac{4R_1^2}{R_1^2}\) \(n = 4\) So, the minimum possible value of \(n\) is 4, which corresponds to option (A).

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Key Concepts

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

Ohm's Law
Ohm's Law is a fundamental principle in electronics and physics. It describes the relationship between voltage, current, and resistance in an electrical circuit. Mathematically, it can be expressed as \( V = IR \), where \( V \) represents the voltage across the resistor, \( I \) is the current flowing through the resistor, and \( R \) is the resistance of the resistor itself.
Understanding Ohm's Law is crucial because it provides the basis for analyzing and determining the values of these three key electrical quantities. In any given resistor, increasing the voltage will lead to an increase in current, provided resistance remains constant. Conversely, increasing resistance while maintaining constant voltage will decrease the current. This concept is used extensively in circuit design and analysis.
  • Voltage (\(V\)) is measured in volts.
  • Current (\(I\)) is measured in amperes.
  • Resistance (\(R\)) is measured in ohms.
By using this law, one can determine any of the three quantities if the other two are known. Ohm's Law is foundational for understanding more complex electrical concepts like resistors in series and parallel combinations.
Electrical Resistivity
Electrical resistivity is a material property that quantifies how strongly a given material opposes the flow of electric current. This property is dependent on several factors including the material itself, its temperature, and its geometric structure. Resistivity is denoted by the symbol \( \rho \) and is typically measured in ohm-meters (\( \Omega \cdot m \)).
Resistivity helps us understand why some materials conduct electricity better than others. Conductors, like copper, have low resistivity, which means they allow electric current to pass through them easily. Insulators, like rubber, have high resistivity, making them good barriers to electric current.
  • Low resistivity implies good conductance.
  • High resistivity implies poor conductance.
  • Resistivity changes with temperature; for most conductors, it increases with rising temperature.
In practical applications, the resistance of a wire (or any material) can be calculated if the material's resistivity, length, and cross-sectional area are known, using the formula: \( R = \frac{\rho L}{A} \). This understanding is crucial when selecting materials for electrical wiring and components in a circuit.
Circuit Analysis
Circuit analysis involves the process of determining the electrical properties of a circuit - mainly the voltage, current, and resistance - within an electrical network. It incorporates a variety of techniques to evaluate complex circuits that consist of series and parallel components.
When analyzing circuits, it's essential to understand the difference between series and parallel resistor arrangements. In a series circuit, resistors are connected end-to-end, so the same current flows through each resistor, but the voltage across each can differ. The total resistance in a series is calculated by simply adding individual resistances: \( R_{\text{total}} = R_1 + R_2 + ... + R_n \).

In parallel circuits, resistors are connected across the same two points, thus sharing the same voltage across each resistor while the current through each can vary. The total resistance in a parallel arrangement is found via the formula: \( \frac{1}{R_{\text{total}}} = \frac{1}{R_1} + \frac{1}{R_2} + ... + \frac{1}{R_n} \).
  • Series circuits add resistances linearly.
  • Parallel circuits result in lesser resistance than the smallest individual resistor.
Understanding series and parallel combinations is crucial for effective circuit design and prediction of circuit behavior. This knowledge allows for proper planning to ensure circuits operate efficiently and safely.

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Most popular questions from this chapter

The resistance of the wire made of silver at \(27^{\circ} \mathrm{C}\) temperature is equal to \(2.1 \Omega\) while at \(100^{\circ} \mathrm{C}\) it is \(2.7 \Omega\) calculate the temperature coefficient of the resistivity of silver. Take the reference temperature equal to \(20^{\circ} \mathrm{C}\) (A) \(4.02 \times 10^{-3}{ }^{\circ} \mathrm{C}^{-1}\) (B) \(0.402 \times 10^{-3}{ }^{\circ} \mathrm{C}^{-1}\) (C) \(40.2 \times 10^{-4}{ }^{\circ} \mathrm{C}^{-1}\) (D) \(4.02 \times 10^{-4}{ }^{\circ} \mathrm{C}^{-1}\)

\(2 \mathrm{~A}\) current is obtained when a \(2 \Omega\) resistor is connected with battery having \(r \Omega\) as internal resistance \(0.5 \mathrm{~A}\) current is obtained if the above battery is connected to \(9 \Omega\) resistor. Calculate the internal resistance of the battery. (A) \(0.5 \Omega\) (B) \((1 / 3) \Omega\) (C) \((1 / 4) \Omega\) (D) \(1 \Omega\)

Length of a wire of resistance \(R \Omega\) is increased to 10 times, so its resistance becomes \(1000 \Omega\), therefore \(R=\ldots .\) (The volume of the wire remains same during increase in length) (A) \(0.01 \Omega\) (B) \(0.1 \Omega\) (C) \(1 \Omega\) (D) \(10 \Omega\)

A Potentiometer wire, \(10 \mathrm{~m}\) long, has a resistance of \(40 \Omega\). It is connected in series with a resistance box and a \(2 \mathrm{v}\) storage cell If the potential gradient along the wire is \(0.1 \mathrm{~m} \mathrm{v} / \mathrm{cm}\), the resistance unplugged in the box is. (A) \(260 \Omega\) (B) \(760 \Omega\) (C) \(960 \Omega\) (D) \(1060 \Omega\)

The potential difference through the \(3 \Omega\) resistor shown in fig is.... (A) Zero (B) \(1 \mathrm{~V}\) (C) \(3.5 \mathrm{~V}\) (D) \(7 \mathrm{~V}\)

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