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Chapter 12: Problem 1794

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) \(1060 \Omega\)

Short Answer

Expert verified
The resistance unplugged in the resistance box is \(760 \Omega\), which corresponds to option (B).

Step by step solution

01

1. Understand the given information and identify what needs to be found

We are given the following information: 1. Length of the potentiometer wire (L): \(10 \mathrm{~m}\) 2. Resistance of the potentiometer wire (R): \(40 \Omega\) 3. Voltage of the storage cell (V): \(2 \mathrm{~V}\) 4. Potential gradient along the wire (K): \(0.1 \mathrm{~mV/cm}\) We need to find the resistance unplugged in the resistance box (Rb).
02

2. Convert potential gradient to volts per meter

The potential gradient is given in millivolts per centimeter. Let's first convert it to volts per meter. 1 mV = \(10^{-3}\) V and 1 cm = \(10^{-2}\) m So, \(K = 0.1 \cdot 10^{-3} \cdot 10^{-2} = 0.1 \cdot 10^{-1} \mathrm{V/m}\)
03

3. Calculate the potential difference across the wire

Potential difference across the wire (Vw) can be calculated using the potential gradient (K) and the length of the wire (L): \[Vw = K \cdot L = 0.1 \cdot 10^{-1} \cdot 10 = 0.1 \mathrm{V}\]
04

4. Calculate the potential difference across the resistance box

Because the resistance box is connected in series with the wire, V = Vw + Vrb, where Vrb is the potential difference across the resistance box. We can solve for Vrb: \[Vrb = V - Vw = 2 - 0.1 = 1.9 \mathrm{V}\]
05

5. Use Ohm's Law to find the resistance in the resistance box

Now, we can use Ohm's Law to find the resistance in the resistance box. Ohm's Law states that V = I * R, where V is the potential difference, I is the current, and R is the resistance. We know the potential difference across the wire, Vw, and its resistance, R. So, we can find the current in the circuit: \[I = \frac{Vw}{R} = \frac{0.1}{40} = \frac{1}{400} \mathrm{A}\] Now, we can use the current and the potential difference across the resistance box to find the resistance unplugged in the box (Rb): \[Rb = \frac{Vrb}{I} = \frac{1.9}{\frac{1}{400}} = 760 \Omega\] So, the resistance unplugged in the resistance box is \(760 \Omega\), which corresponds to option (B).

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Most popular questions from this chapter

Formula for current flowing through a wire is \(I=6 t^{2}+4 t-2\) here \(\mathrm{t}\) is in second and I is an ampere. In this wire, what is the quantity of electric charge passing in time interval \(1 \mathrm{sec}\) to 2 sec? (A) \(8 \mathrm{C}\) (B) \(18 \mathrm{C}\) (C) \(20 \mathrm{C}\) (D) \(24 \mathrm{C}\)

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