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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

When a \(40.0-V\) emf device is placed across two resistors in series, a current of \(10.0 \mathrm{~A}\) is flowing in each of the resistors. When the same emf device is placed across the same two resistors in parallel, the current through the emf device is \(50.0 \mathrm{~A}\). What is the magnitude of the larger of the two resistances?

Before bendable tungsten filaments were developed, Thomas Edison used carbon filaments in his light bulbs. Though carbon has a very high melting temperature \(\left(3599^{\circ} \mathrm{C}\right)\) its sublimation rate is high at high temperatures. So carbonfilament bulbs were kept at lower temperatures, thereby rendering them dimmer than later tungsten-based bulbs. A typical carbon-filament bulb requires an average power of \(40 \mathrm{~W}\), when 110 volts is applied across it, and has a filament temperature of \(1800^{\circ} \mathrm{C}\). Carbon, unlike copper, has a negative temperature coefficient of resistivity: \(\alpha=-0.0005^{\circ} \mathrm{C}^{-1}\) Calculate the resistance at room temperature \(\left(20^{\circ} \mathrm{C}\right)\) of this carbon filament.

Three resistors are connected to a power supply with \(V=110 . \mathrm{V}\) as shown in the figure a) Find the potential drop across \(R_{3}\) b) Find the current in \(R_{1}\). c) Find the rate at which thermal energy is dissipated from \(R_{2}\).

A material is said to be ohmic if an electric field, \(\vec{E}\), in the material gives rise to current density \(\vec{J}=\sigma \vec{E},\) where the conductivity, \(\sigma\), is a constant independent of \(\vec{E}\) or \(\vec{J}\). (This is the precise form of Ohm's Law.) Suppose in some material an electric field, \(\vec{E}\), produces current density, \(\vec{J},\) not necessarily related by Ohm's Law; that is, the material may or may not be ohmic. a) Calculate the rate of energy dissipation (sometimes called ohmic heating or joule heating) per unit volume in this material, in terms of \(\vec{E}\) and \(\vec{J}\). b) Express the result of part (a) in terms of \(\vec{E}\) alone and \(\vec{J}\) alone, for \(\vec{E}\) and \(\vec{J}\) related via Ohm's Law, that is, in an ohmic material with conductivity \(\sigma\) or resistivity \(\rho .\)

The Stanford Linear Accelerator accelerated a beam consisting of \(2.0 \cdot 10^{14}\) electrons per second through a potential difference of \(2.0 \cdot 10^{10} \mathrm{~V}\) a) Calculate the current in the beam. b) Calculate the power of the beam. c) Calculate the effective ohmic resistance of the accelerator.
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