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Chapter 10: Problem 12

Suppose you measure the terminal voltage of a \(1.585-\mathrm{V}\) alkaline cell having an internal resistance of \(0.100 \Omega\) by placing a \(1.00-\mathrm{k} \Omega\) voltmeter across its terminals. (See Figure.) (a) What current flows? (b) Find the terminal voltage. (c) To see how close the measured terminal voltage is to the emf, calculate their ratio

Short Answer

Expert verified
The current flowing is found to be approximately 1.58 mA, the terminal voltage is approximately 1.585 V, and the ratio of terminal voltage to EMF is approximately 0.9997.

Step by step solution

01

Calculate the Current Flowing Through the Circuit

Using Ohm's Law, we calculate the current using the formula \(I = \frac{V}{R_{total}}\), where \(V\) is the EMF of the cell, and \(R_{total}\) is the total resistance, which is the sum of the internal resistance of the cell and the resistance of the voltmeter. Calculate the total resistance as \(R_{total} = R_{internal} + R_{voltmeter} = 0.100 \Omega + 1.00 \times 10^3 \Omega\).
02

Use Ohm's Law to Find Terminal Voltage

With the current known, calculate the terminal voltage using the formula \(V_{terminal} = V - I \times R_{internal}\). The actual voltage measured by the voltmeter will be lower than the EMF of the cell due to the voltage drop across the internal resistance.
03

Calculate the Ratio of Terminal Voltage to EMF

The ratio indicates how close the measured terminal voltage is to the EMF of the battery, calculated by dividing the terminal voltage by the EMF: \(\text{Ratio} = \frac{V_{terminal}}{V}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Electromotive Force (EMF)
Electromotive force (EMF) is a term used to describe the energy provided by a battery or a generator per Coulomb of charge. It is the driving force that pushes electrons through a circuit, which is why it is often thought of as 'electrical pressure'. Formally, it's measured in volts (V) and, despite the name, it's not actually a force but a potential difference 鈥� a concept that can be confusing for students just beginning to study electronics.

When a circuit is open, the EMF is the maximum voltage a power source can provide. However, as soon as you connect a load, like in the exercise where a voltmeter is connected across a cell, the EMF is no longer the only factor in play as the internal resistance of the battery and resistance of the connected circuits also come into the equation.
The Role of Internal Resistance
Internal resistance is an often-overlooked aspect of electrical circuits which can dramatically affect their performance. Simply put, it's the resistance within the battery or power source itself. It is the reason why the actual voltage available to an external circuit (terminal voltage) is less than the EMF of the battery.

All real batteries have some amount of this resistance, and it causes the battery to lose some of its energy as heat when it provides current. In the given exercise, this concept is crucial because the internal resistance, together with the resistance of the voltmeter, reduces the current flowing through the circuit. We see this when we use Ohm鈥檚 Law to calculate the total resistance and then find the current, taking into account both the resistance of the voltmeter and the internal resistance.
Terminal Voltage Explained
Terminal voltage is the voltage output of a battery that is available to a circuit. It is the voltage you would measure when you connect a voltmeter across the terminals of a battery when it is delivering current. Terminal voltage is less than the EMF of the battery because some voltage is lost over the internal resistance of the power source. It's an essential concept for practical electrical engineering, as it determines how much power you can actually use from a battery.

In the exercise, you'll see that after calculating the current, the terminal voltage is found by taking into account the voltage drop across the internal resistance. Taking the time to understand the relationship between EMF, internal resistance, and terminal voltage will give students a much better grasp of how batteries behave in real-world circuits.

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Most popular questions from this chapter

A car battery with a 12-V emf and an internal resistance of \(0.050 \Omega\) is being charged with a current of \(60 \mathrm{~A}\). Note that in this process the battery is being charged. (a) What is the potential difference across its terminals? (b) At what rate is thermal energy being dissipated in the battery? (c) At what rate is electric energy being converted to chemical energy? (d) What are the answers to (a) and (b) when the battery is used to supply \(60 \mathrm{~A}\) to the starter motor?

(a) Given a 48.0-V battery and \(24.0-\Omega\) and \(96.0-\Omega\) resistors, find the current and power for each when connected in series. (b) Repeat when the resistances are in parallel.

While exercising in a fitness center, a man lies face down on a bench and lifts a weight with one lower leg by contacting the muscles in the back of the upper leg. (a) Find the angular acceleration produced given the mass lifted is \(10.0 \mathrm{kg}\) at a distance of \(28.0 \mathrm{cm}\) from the knee joint, the moment of inertia of the lower leg is \(0.900 \mathrm{kg} \cdot \mathrm{m}^{2}\), the muscle force is 1500 N, and its effective perpendicular lever arm is \(3.00 \mathrm{cm}\). (b) How much work is done if the leg rotates through an angle of \(20.0^{\circ}\) with a constant force exerted by the muscle?

An automobile starter motor has an equivalent resistance of \(0.0500 \Omega\) and is supplied by a \(12.0-\mathrm{V}\) battery with a \(0.0100-\Omega\) internal resistance. (a) What is the current to the motor? (b) What voltage is applied to it? (c) What power is supplied to the motor? (d) Repeat these calculations for when the battery connections are corroded and add \(0.0900 \Omega\) to the circuit. (Significant problems are caused by even small amounts of unwanted resistance in low-voltage, high-current applications.)

The timing device in an automobile's intermittent wiper system is based on an \(R C\) time constant and utilizes a \(0.500-\mu \mathrm{F}\) capacitor and a variable resistor. Over what range must \(R\) be made to vary to achieve time constants from \(2.00\) to \(15.0 \mathrm{~s}\) ?
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