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Chapter 19: Problem 50

An ammeter has resistance \(R_{0}\) and range \(I\). What resistance should be connected in parallel with it to increase its range by \(n I\) ? (a) \(R_{0} /(n-1)\) (b) \(R_{0} /(n+1)\) (c) \(R_{0} / n\) (d) None of these

Short Answer

Expert verified
(c) \( R_{0} / n \)

Step by step solution

01

Analyze the Required Increase in Current Range

The problem states that the current range of the ammeter should be increased to \(nI\). This means the total range should be \(I + nI = (n+1)I\). We need to find the resistance needed in parallel to achieve this increase.
02

Understand the Ammeter Characteristics

The ammeter's internal resistance is given as \(R_{0}\) and it is originally capable of handling a maximum current \(I\). The voltage across the ammeter can be expressed as \(V = I imes R_{0}\).
03

Calculate New Ammeter Current with Parallel Resistance

With a parallel resistance \(R_{p}\), the new total current range is \(I + I_p = (n+1)I\), where \(I_p\) is the current flowing through \(R_{p}\). Using the voltage across the ammeter, \(I_p = \, \frac{V}{R_{p}} = \, \frac{IR_{0}}{R_{p}}\).
04

Set Up Equation for New Total Current

Since \(I + I_p = (n+1)I\), substitute \(I_p = \, \frac{IR_{0}}{R_{p}}\) in: \( I + \frac{IR_{0}}{R_{p}} = (n+1)I\). Simplify this to find: \( \frac{IR_{0}}{R_{p}} = nI\).
05

Solve for the Parallel Resistance

By solving \( \frac{IR_{0}}{R_{p}} = nI\) for \(R_{p}\), we find: \( R_{p} = \frac{R_{0}}{n} \). Hence, the resistance needed in parallel to achieve the required range is \( R_{0}/n\).
06

Verify the Solution Choice

Compare the derived expression \( R_{0}/n \) with the given options: (a) \(R_{0} /(n-1)\), (b) \(R_{0} /(n+1)\), (c) \(R_{0} / n\), (d) None of these. The correct and matching answer is (c) \( R_{0}/n \).

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Key Concepts

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

Parallel Resistance
Parallel resistance is a concept where two or more resistors are connected in such a way that the same voltage is across each of them, but the total current flowing through the system can vary. When resistors are in parallel, the voltage drop across each is the same, but they share the current flow.
This setup can be incredibly useful in electronic circuits, especially when trying to adjust or control the current passing through a component like the ammeter. For example, if you want to increase the range of an ammeter, you can connect a parallel resistor to aid in distributing the total current.
In the case of our exercise, this added resistor helps to partition the current between the ammeter and the parallel path, allowing the ammeter to measure higher current ranges without being damaged.
Ohm's Law
Ohm's Law is fundamental in understanding electrical circuits. It states that the current flow through a conductor between two points is directly proportional to the voltage across the two points and inversely proportional to the resistance. Mathematically, Ohm's Law is given by the expression \( V = I \times R \).
In the ammeter context of the exercise, Ohm's Law helps us make calculations to understand how changing the resistance in the circuit influences the current flow. By adding a parallel resistor, we're essentially applying Ohm's Law to figure out how much current will flow through each path.
Understanding Ohm鈥檚 Law can make it easier to predict how changes in a circuit鈥攊n this case increasing the ammeter's range鈥攚ill affect current and voltage relationships. It tells us why adding a resistor in parallel allows for more current to run through the device.
Current Range Expansion
Expanding the current range of an ammeter is crucial when you want to measure currents that exceed the device's initial capabilities without damaging it. This is achieved by implementing additional components like resistors in parallel to aid in rerouting part of the total current.
In our problem, we're looking at increasing the current range from \( I \) to \( (n+1)I \). To do this, we add a parallel resistor, which helps share the additional current, allowing the ammeter to safely measure higher values.
The solution involves calculating the suitable resistance that when placed in parallel, increases the current capacity of the ammeter without exceeding its voltage rating. By using the determined parallel resistance \( R_{0}/n \), the ammeter's range is effectively expanded to handle the desired increased current range.

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Most popular questions from this chapter

A steady current \(i\) flows in a small square loop of wire of side \(l\) in a horizontal plane. The loop is now folded about its middle such that half of it lies in a vertical plane. Let \(M_{1}\) and \(M_{2}\) respectively denote the magnetic moments due to current loop before and after folding. Then (a) \(M_{2}=0\) (b) \(M_{1}\) and \(M_{2}\) are in the same direction (c) \(M_{1} / M_{2}=\sqrt{2}\) (d) \(M_{1} / M_{2}=1 / \sqrt{2}\)

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An infinitely long wire carrying current \(i\) is along \(Y\)-axis such that its one end is at point \((0, b)\) while the wire extends upto \(\infty .\) The magnitude of magnetic field strength at point \(P(a, 0)\) is (a) \(\frac{\mu_{0} i}{4 \pi a}\left(1+\frac{b}{\sqrt{a^{2}+b^{2}}}\right)\) (b) \(\frac{\mu_{0} i}{4 \pi a}\left(1-\frac{b}{\sqrt{a^{2}+b^{2}}}\right)\) (c) \(\frac{\mu_{0} i}{4 \pi a}\left(1-\frac{a}{\sqrt{a^{2}+b^{2}}}\right)\) (d) \(\frac{\mu_{0} i}{4 \pi a}\left(1+\frac{a}{\sqrt{a^{2}+b^{2}}}\right)\)
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