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Chapter 26: Problem 12

In these Problems neglect the internal resistance of a battery unless the Problem refers to it. (II) Eight identical bulbs are connected in series across a 110-V line. ( \(a\) ) What is the voltage across each bulb? (b) If the current is \(0.42 \mathrm{~A}\), what is the resistance of each bulb, and what is the power dissipated in each?

Short Answer

Expert verified
Each bulb has 13.75 V, a resistance of 32.74 ohms, and dissipates 5.78 W of power.

Step by step solution

01

Understand the Series Circuit

In a series circuit, the same current flows through each component, and the total voltage is distributed across all components. Here, we have eight identical bulbs connected in series.
02

Calculate Voltage Across Each Bulb

To find the voltage across each bulb, divide the total voltage of the line by the number of bulbs. Since the total voltage is 110 V and there are 8 bulbs, the voltage across each bulb is:\[V_{bulb} = \frac{110 \text{ V}}{8} = 13.75 \text{ V}\]
03

Identify Current and Apply Ohm's Law

The problem states the current in the circuit is 0.42 A. Using Ohm鈥檚 Law \( V = IR \), where \( I \) is the current and \( R \) is the resistance, we can calculate the resistance of each bulb.
04

Calculate Resistance of Each Bulb

The voltage across each bulb is 13.75 V and the current is 0.42 A:\[R_{bulb} = \frac{V_{bulb}}{I} = \frac{13.75 \text{ V}}{0.42 \text{ A}} \approx 32.74 \text{ ohms}\]
05

Calculate Power Dissipated in Each Bulb

The power dissipated in a bulb can be found using the formula \( P = IV \), where \( P \) is power, \( I \) is current, and \( V \) is voltage. For each bulb:\[P_{bulb} = 0.42 \text{ A} \times 13.75 \text{ V} \approx 5.78 \text{ W}\]

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Key Concepts

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

Ohm's Law
Ohm's Law is a fundamental principle used to understand electrical circuits. It states that the voltage (\(V\)) across a resistor is proportional to the current (\(I\)) flowing through it, with the proportionality constant being the resistance (\(R\)). Mathematically, Ohm's Law is expressed as \( V = IR \).

In the context of our exercise, this equation allows us to calculate the resistance of each bulb in a series circuit. By knowing the voltage across each bulb (13.75 V) and the current flowing through the circuit (0.42 A), we can rearrange Ohm's Law to solve for resistance: \( R = \frac{V}{I} = \frac{13.75 \mathrm{~V}}{0.42 \mathrm{~A}} \approx 32.74 \text{ ohms} \).

This simple formula is essential for analyzing and designing circuits, ensuring that components operate within their intended parameters.
Voltage Distribution
In a series circuit, the total voltage supplied by the battery is divided among the connected components. This phenomenon is known as voltage distribution. The voltage drop across each component, such as a bulb, depends on its resistance relative to the total resistance of the circuit. In our exercise, the total voltage of 110 V is distributed evenly across eight identical bulbs.

This means each bulb receives an equal share of the total voltage. With eight bulbs, the voltage across each is calculated as \( V_{bulb} = \frac{110 \text{ V}}{8} = 13.75 \text{ V} \).

It is crucial to understand this concept when designing circuits, as it ensures that each component receives the appropriate voltage to function correctly. Voltage distribution principles also apply to more complex circuits with different kinds and numbers of components.
Power Dissipation
Power dissipation refers to the conversion of electrical energy into thermal energy (or heat) in a circuit component, such as a bulb. The power dissipated by a component is calculated using the formula \( P = IV \), where \( P \) is power, \( I \) is current, and \( V \) is voltage across the component.

In our series circuit problem, each bulb dissipates power through the current of 0.42 A and voltage of 13.75 V. Calculating the power for one bulb: \( P_{bulb} = 0.42 \text{ A} \times 13.75 \text{ V} \approx 5.78 \text{ W} \).

This indicates how much energy is converted into heat by each bulb every second. Understanding power dissipation is essential for ensuring that components do not overheat and can help in designing circuits to avoid energy waste.

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Most popular questions from this chapter

(II) A parallel-plate capacitor is filled with a dielectric of dielectric constant \(K\) and high resistivity \(\rho\) (it conducts very slightly). This capacitor can be modeled as a pure capacitance \(C\) in parallel with a resistance \(R\). Assume a battery places a charge \(+Q\) and \(-Q\) on the capacitor's opposing plates and is then disconnected. Show that the capacitor discharges with a time constant \(\tau=K \varepsilon_{0} \rho\) (known as the dielectric relaxation time). Evaluate \(\tau\) if the dielectric is glass with \(\rho=1.0 \times 10^{12} \Omega \cdot \mathrm{m}\) and \(K=5.0\).

(II) A close inspection of an electric circuit reveals that a \(480-\Omega\) resistor was inadvertently soldered in the place where a \(370-\Omega\) resistor is needed. How can this be fixed without removing anything from the existing circuit?

In these Problems neglect the internal resistance of a battery unless the Problem refers to it. (I) Suppose that you have a \(680-\Omega,\) a \(720-\Omega,\) and a \(1.20-\mathrm{k} \Omega\) resistor. What is \((a)\) the maximum, and \((b)\) the minimum resistance you can obtain by combining these?

IThe Problems in this Section are ranked I, II, or III according to estimated difficulty, with (I) Problems being easiest. Level (III) Problems are meant mainly as a challenge for the best students, for "extra credit." The Problems are arranged by Sections, meaning that the reader should have read up to and including that Section, but this Chapter also has a group of General Problems that are not arranged by Section and not ranked.] $$ \begin{array}{l}{\text { (I) Calculate the terminal voltage for a battery with an }} \\ {\text { internal resistance of } 0.900 \Omega \text { and an emf of } 6.00 \mathrm{V} \text { when the }} \\ {\text { battery is connected in series with }(a) \text { an } 81.0-\Omega \text { resistor, and }} \\ {\text { (b) an } 810-\Omega \text { resistor. }}\end{array} $$

(II) A galvanometer has an internal resistance of \(32 \Omega\) and deflects full scale for a \(55-\mu\) A current. Describe how to use this galvanometer to make \((a)\) an ammeter to read currents up to \(25 \mathrm{~A}\), and \((b)\) a voltmeter to give a full scale deflection $$\text { of } 250 \mathrm{~V} \text { . }$$
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