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Chapter 12: Problem 1788

For a cell of e.m.f \(2 \mathrm{~V}\), a balance is obtained for $50 \mathrm{~cm}\( of the potentiometer wire If the cell is shunted by a \)2 \Omega$ resistor and the balance is obtained across \(40 \mathrm{~cm}\) of the wire, then the internal resistance of the cell is. (A) \(1 \Omega\) (B) \(0.5 \Omega\) (C) \(1.2 \Omega\) (D) \(2.5 \Omega\)

Short Answer

Expert verified
The internal resistance of the cell is \(0.5 \Omega\), which corresponds to option (B).

Step by step solution

01

Understand the standard conditions

In the standard potentiometer setup (without shunting), we have: - Cell EMF: \(E = 2V\) - Length of wire: \(L_1 = 50cm\) The potential per unit length (V/cm) is given by the formula: \(K = E / L_1\). Using this information, we can calculate K.
02

Calculate K for the standard condition

To calculate the potential per unit length K, we use the formula: \[K = E / L_1\] Plugging the values for the EMF and the length of the wire, we get: \[K = 2V / 50cm = 0.04 V/cm\]
03

Understand the shunt conditions

Now, let's look at the shunt conditions. The cell is shunted with a 2-ohm resistor, and the balance is obtained across 40cm of the wire. In this case, we have: - Length of wire: \(L_2 = 40cm\) - Shunt resistor: \(R_s = 2\Omega\) From this, we can calculate the voltage across the 40cm wire when a shunt is placed across the cell.
04

Calculate the voltage across the wire during shunt

Recalling the relationship of potential per unit length, we have: \[V = K \cdot L_2\] \[V = 0.04 V/cm \cdot 40 cm\] \[V = 1.6 V\] The EMF of the cell remains the same, so it must be shared between the internal resistance and the effective external resistance.
05

Set up an equation to solve for the internal resistance

Let \(r\) be the internal resistance of the cell. Using the voltage divider rule: \[V = E \cdot \frac{R_s}{R_s + r}\] Plugging in the values, we can set up an equation to solve for the internal resistance: \[1.6 V = 2 V \cdot \frac{2\Omega}{2\Omega + r}\]
06

Solve for the internal resistance \(r\)

To solve for the internal resistance \(r\), First, we can cross multiply the fraction and simplify the equation: \[1.6 \cdot (2\Omega + r) = 4 \Omega\] Expand and simplify: \[3.2 \Omega + 1.6 r = 4 \Omega\] Then, subtract 3.2 from both sides: \[1.6 r = 0.8 \Omega\] Now, divide both sides by 1.6: \[r = 0.5 \Omega\] The internal resistance of the cell is 0.5 ohms, which corresponds to option (B).

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Most popular questions from this chapter

A bulb of \(300 \mathrm{~W}\) and \(220 \mathrm{~V}\) is connected with a source of \(110 \mathrm{~V}\). What is the \(\%\) decrease in power? (A) \(100 . \%\) (B) \(75 \%\) (C) \(70 \%\) (D) \(25 \%\)

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