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Chapter 18: Problem 117

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 \((0.1 \mathrm{mV} / \mathrm{cm})\), the resistance unplugged in the box is (a) \(260 \Omega\) (b) \(760 \Omega\) (c) \(960 \Omega\) (d) \(1060 \Omega\)

Short Answer

Expert verified
The resistance unplugged in the box is \(760 \Omega\) (Option b).

Step by step solution

01

Understanding Potential Gradient

The potential gradient is given as \(0.1 \text{ mV/cm}\). To convert it to volts per meter, note that 1 mV = 0.001 V and 1 cm = 0.01 m. Therefore, the potential gradient is \(0.1 \text{ mV/cm} = 0.01 \text{ V/m}\).
02

Calculating Total Voltage Across Wire

The potential drop (V) across the full length of the wire can be calculated using the formula \(V = k \times L\), where \(k = 0.01 \text{ V/m}\) and \(L = 10 \text{ m}\). Hence, \(V = 0.01 \times 10 = 0.1 \text{ V}\).
03

Using Ohm's Law to Determine Total Resistance

Using Ohm's Law, where \(I\) is the current through the circuit, \(R_{total}\) is the total resistance in the circuit, and \(V \text{ is the voltage of the battery (2 V)}\), we can say \(2 = I \times R_{total}\). Since \(V = I \times 40\) for the wire, \(I = \frac{0.1}{40}\).
04

Calculating Total Circuit Resistance

The voltage drop across the wire is \(0.1 \text{ V}\). Thus, \(I = \frac{0.1}{40} = 0.0025\text{ A}\). Substituting this current into the total voltage equation, \(2 = 0.0025 \times R_{total}\), we find \(R_{total} = 800 \Omega\).
05

Determining Resistance in the Box

The total resistance in the circuit, \(R_{total}\), is the sum of the wire's resistance (\(40 \Omega\)) and the resistance in the box. Therefore, \(R_{box} = R_{total} - 40 = 800 - 40 = 760 \Omega\).

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Key Concepts

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

Potential Gradient
The potential gradient is a measure of how the potential (voltage) changes along a conductor. It can be thought of as the rate at which the potential decreases per unit length of the wire. This is crucial for understanding how voltage is distributed in a circuit.
In the given exercise, the potential gradient is specified as 0.1 mV/cm. To work with standard units in physics, it's important to convert this to volts per meter (V/m).
We know that:
  • 1 mV = 0.001 V
  • 1 cm = 0.01 m
Using these conversions, 0.1 mV/cm is equivalent to 0.01 V/m.
This means that for every meter of the wire, the voltage drops by 0.01 volts. This concept helps us understand how the energy is distributed along the wire's length.
Ohm's Law
Ohm's Law is one of the fundamental principles in electronics, relating voltage (V), current (I), and resistance (R) in an electric circuit. The formula is expressed as: \[ V = I \times R \] This equation tells us how much current will flow through a given resistance when a certain voltage is applied.
Applying Ohm鈥檚 Law to this exercise:
  • The wire has a resistance of 40 惟 and a potential drop of 0.1 V along its length.
  • Using Ohm's Law, the current through the wire can be calculated as: \[ I = \frac{0.1}{40} = 0.0025 \, \text{A} \]
Knowing the current is essential for further calculations in determining the total resistance in the circuit.
Resistance Calculation
Resistance is a key factor in controlling and managing the flow of electric current. It is the property of a material that resists the flow of current through it.
In the exercise, we determine the resistance in an unknown resistor box. This is achieved by first calculating the total resistance in the circuit, which includes the wire and the resistance box.
Given:
  • The total resistance in the circuit (denoted as \( R_{total} \)) is 800 惟.
  • The wire's resistance is 40 惟.
The resistance of the box \( (R_{box}) \) can be found using: \[ R_{box} = R_{total} - R_{wire} \] Substituting the values, we get: \[ R_{box} = 800 - 40 = 760 \, \Omega \] This calculation shows how resistance calculations help in understanding how different portions of a circuit contribute to total resistance.
Electric Circuits
An electric circuit is a closed path that allows the flow of electric charge. It typically comprises a power source, components such as resistors, and connecting conductors like wires.
In our study, the electric circuit is formed by:
  • A potentiometer wire with known resistance.
  • An unknown resistor box.
  • A power source (2 V storage cell).
This setup is used to measure the potential gradient and calculate the total resistance in the circuit.
Understanding electric circuits involves recognizing how these elements interact:
  • The power source provides the voltage that pushes current through the circuit.
  • The resistors regulate this current flow by offering resistance.
Through the circuit analysis, one can determine meaningful parameters like current value, total resistance, and potential drops across different components. Such insights are critical in designing and assessing electrical systems.

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Most popular questions from this chapter

A teacher asked a student to connect \(N\) cells each of \(\operatorname{emf} e\) in series to get a total emf of \(\mathrm{Ne}\). While connecting, the student, by mistake, reversed the polarity of \(n\) cells. The total emf of the resulting series combination is (a) \(e\left(N-\frac{n}{2}\right)\) (b) \(e(N-n)\) (c) \(e(N-2 n)\) (d) \(e N\)

The length of a potentiometer wire is \(l\). A cell of emf \(E\) is balanced at a length \(l / 3\) from the positive end of the wire. If the length of the wire is increased by \(1 / 2\), at what distance will the same cell gives a balance point (a) \(2 \mathrm{ll} / 3\) (b) \(l / 2\) (c) \(l / 6\) (d) \(4 l / 3\)

When a potential difference is applied across a conductor, the free electrons in the conductor are set into motion. Two velocities are associated with the moving electrons, the drift velocity and average velocity. The fact is that the two are (a) entirely different (b) same (c) same in some conductors but different in others (d) none of the above

An aluminium (Al) rod with area of cross-section \(4 \times 10^{-6} \mathrm{~m}^{2}\) has a current of \(5 \mathrm{~A}\), flowing through it. Find the drift velocity of electron in the rod. Density of \(\mathrm{Al}=2.7 \times 10^{3} \mathrm{~kg}^{3} / \mathrm{m}^{3}\) and atomic wt. \(=27\). Assume that each Al atom provides one electron (a) \(8.6 \times 10^{-4} \mathrm{~m} / \mathrm{s}\) (b) \(6.2 \times 10^{-4} \mathrm{~m} / \mathrm{s}\) (c) \(2.8 \times 10^{-2} \mathrm{~m} / \mathrm{s}\) (d) \(0.13 \times 10^{-3} \mathrm{~m} / \mathrm{s}\)

The maximum current that flows in the fuse wire, before it blows out, varies with the radius \(r\) as (a) \(r^{3 / 2}\) (b) \(r\) (c) \(r^{2 / 3}\) (d) \(r^{\mathrm{L} / 2}\)
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