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Chapter 25: Problem 43

A battery has a potential difference of \(14.50 \mathrm{~V}\) when it is not connected in a circuit. When a \(17.91-\Omega\) resistor is connected across the battery, the potential difference of the battery drops to \(12.68 \mathrm{~V}\). What is the internal resistance of the battery?

Short Answer

Expert verified
Answer: The approximate internal resistance of the battery is 1.67 Ω.

Step by step solution

01

Understand the problem

First, we need to understand the concept involved in this problem. A battery has an internal resistance, which causes the potential difference (voltage) across the battery to change depending on the electrical load connected to it. In this case, we are given the open-circuit voltage (no load connected) and the voltage when a load (resistor) is connected across it. We need to find the internal resistance of the battery using these values.
02

Apply Ohm's law

To solve this problem, we will apply Ohm's law, which states that the potential difference across a resistor is equal to the product of the current passing through it and its resistance, i.e., \(V=IR\). We will write an equation for the potential difference across the external resistor and the potential difference across the internal resistance of the battery when the load is connected.
03

Write the equations

When the load is connected, the potential difference across the battery is 12.68 V. Therefore, \(V_\text{external} + V_r = 12.68\) Where: - \(V_\text{external}\) is the potential difference across the internal resistance of the battery, - \(V_r\) is the potential difference across the external resistor. We can rewrite this equation using Ohm's law as: \(I r_\text{int} + IR = 12.68\) Where: - \(I\) is the current passing through the circuit, - \(r_\text{int}\) is the internal resistance of the battery, - \(R\) is the resistance of the external resistor (17.91 Ohms).
04

Calculate the current

The potential difference across the external resistor is 12.68 V, and its resistance is 17.91 Ohms. We can find the current passing through the external resistor using Ohm's law: \(I = \frac{V_r}{R} = \frac{12.68}{17.91} = 0.7076 \mathrm{~A}\)
05

Find the internal resistance

Now, we can substitute the current value in the equation: \(0.7076 r_\text{int} + 0.7076 \times 17.91 = 12.68\) Let's solve for \(r_\text{int}\): \(r_\text{int} = \frac{12.68 - 0.7076 \times 17.91}{0.7076} = 1.67 \mathrm{~\Omega}\) (approximately) The internal resistance of the battery is approximately \(1.67 \mathrm{~\Omega}\).

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Most popular questions from this chapter

The most common material used for sandpaper, silicon carbide, is also widely used in electrical applications. One common device is a tubular resistor made of a special grade of silicon carbide called carborundum. A particular carborundum resistor (see the figure) consists of a thick-walled cylindrical shell (a pipe) of inner radius \(a=\) \(1.50 \mathrm{~cm},\) outer radius \(b=2.50 \mathrm{~cm},\) and length \(L=60.0 \mathrm{~cm} .\) The resistance of this carborundum resistor at \(20 .{ }^{\circ} \mathrm{C}\) is \(1.00 \Omega\). a) Calculate the resistivity of carborundum at room temperature. Compare this to the resistivities of the most commonly used conductors (copper, aluminum, and silver). b) Carborundum has a high temperature coefficient of resistivity: \(\alpha=2.14 \cdot 10^{-3} \mathrm{~K}^{-1} .\) If, in a particular application, the carborundum resistor heats up to \(300 .{ }^{\circ} \mathrm{C},\) what is the percentage change in its resistance between room temperature \(\left(20 .{ }^{\circ} \mathrm{C}\right)\) and this operating temperature?

A potential difference of \(12.0 \mathrm{~V}\) is applied across a wire of cross-sectional area \(4.50 \mathrm{~mm}^{2}\) and length \(1000 . \mathrm{km} .\) The current passing through the wire is \(3.20 \cdot 10^{-3} \mathrm{~A}\). a) What is the resistance of the wire? b) What type of wire is this?

A rectangular wafer of pure silicon, with resistivity \(\rho=2300 \Omega \mathrm{m},\) measures \(2.00 \mathrm{~cm}\) by \(3.00 \mathrm{~cm}\) by \(0.010 \mathrm{~cm}\) Find the maximum resistance of this rectangular wafer between any two faces.

Two resistors with resistances \(R_{1}\) and \(R_{2}\) are connected in parallel. Demonstrate that, no matter what the actual values of \(R_{1}\) and \(R_{2}\) are, the equivalent resistance is always less than the smaller of the two resistances.

What is the resistance of a copper wire of length \(l=\) \(10.9 \mathrm{~m}\) and diameter \(d=1.3 \mathrm{~mm} ?\) The resistivity of copper is \(1.72 \cdot 10^{-8} \Omega \mathrm{m}\)
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