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Chapter 25: Problem 45

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 internal resistance of the battery is approximately 3.01 Ω.

Step by step solution

01

Identify the given information

The potential difference when the battery is not connected in a circuit is given as \(V_1 = 14.50 \mathrm{~V}\). When a resistor of resistance \(R = 17.91 \Omega\) is connected in the circuit, the potential difference drops to \(V_2 = 12.68 \mathrm{~V}\). We are asked to find the internal resistance of the battery, which we will denote as \(r\).
02

Determine the total resistance in the circuit

When the resistor is connected across the battery, the total resistance in the circuit is the sum of the external resistor's resistance and the internal resistance of the battery. Mathematically, this can be expressed as: \(R_T = R + r\)
03

Calculate the current in the circuit

We can use Ohm's Law to calculate the current flowing through the circuit when the potential difference across the battery is \(V_2 = 12.68 \mathrm{~V}\). This can be written as: \(I = \frac{V_2}{R_T}\)
04

Calculate the current when the battery is not connected in a circuit

When the battery is not connected in a circuit, we can use Ohm's Law to find the current flowing through the internal resistance. This can be written as: \(I' = \frac{V_1 - V_2}{r}\)
05

Equate the currents and solve for internal resistance

Since the current remains constant throughout the circuit, we can equate the currents found in Step 3 and Step 4, and then solve for the internal resistance \(r\). The equation is as follows: \(\frac{V_2}{R_T} = \frac{V_1 - V_2}{r}\) Substitute the values of \(V_1\), \(V_2\) and \(R\) into this equation and solve for \(r\): \(\frac{12.68}{17.91 + r} = \frac{14.50 - 12.68}{r}\) Now, we can cross-multiply and simplify the equation: \(12.68r = (14.50 - 12.68)(17.91 + r)\) \(12.68r= 1.82(17.91 + r)\) Expand and solve for \(r\): \(12.68r = 32.66 + 1.82r\) \(10.86r = 32.66\) \(r = \frac{32.66}{10.86}\) \(r \approx 3.01 \Omega\)
06

State the final answer

The internal resistance of the battery is approximately \(3.01\,\Omega\).

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Most popular questions from this chapter

Ohm's Law states that the potential difference across a device is equal to a) the current flowing through the device times the resistance of the device. b) the current flowing through the device divided by the resistance of the device. c) the resistance of the device divided by the current flowing through the device. d) the current flowing through the device times the cross-sectional area of the device, e) the current flowing through the device times the length of the device.

A 34-gauge copper wire \(\left(A=0.0201 \mathrm{~mm}^{2}\right),\) with a constant potential difference of \(0.100 \mathrm{~V}\) applied across its \(1.00 \mathrm{~m}\) length at room temperature \(\left(20.0^{\circ} \mathrm{C}\right),\) is cooled to liquid nitrogen temperature \(\left(77 \mathrm{~K}=-196^{\circ} \mathrm{C}\right)\) a) Determine the percentage change in the wire's resistance during the drop in temperature. b) Determine the percentage change in current flowing in the wire. c) Compare the drift speeds of the electrons at the two temperatures.

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 \(20.0^{\circ} \mathrm{C}\) is \(1.00 \mathrm{f}\). 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 \(\left(20.0^{\circ} \mathrm{C}\right)\) and this operating temperature?

When a flashlight bulb with a tungsten filament is lit, the applied potential difference is \(3.991 \mathrm{~V}\) and the temperature of the tungsten filament is \(1.110 \cdot 10^{3}{ }^{\circ} \mathrm{C}\). The resistance of the bulb when it is at room temperature \(\left(20.00^{\circ} \mathrm{C}\right)\) and is not lit is \(1.451 \Omega\). What current does the bulb draw when it is lit?

A voltage spike causes the line voltage in a home to jump rapidly from \(110 . V\) to \(150 . V\). What is the percentage increase in the power output of a \(100 .-\mathrm{W}\) tungsten-filament incandescent light bulb during this spike, assuming that the bulb's resistance remains constant?
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