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Chapter 28: Problem 24

Two resistors, of \(4.00 \Omega\) and \(12.0 \Omega\), are connected in parallel across a 22-V battery having internal resistance \(1.00 \Omega\). Compute \((a)\) the battery current, \((b)\) the current in the \(4.00-\Omega\) resistor, \((c)\) the terminal voltage of the battery, \((d)\) the current in the \(12.0-\Omega\) resistor.

Short Answer

Expert verified
a) 5.50 A; b) 4.125 A; c) 16.5 V; d) 1.375 A.

Step by step solution

01

Calculate Total Resistance in Parallel

For resistors in parallel, the reciprocal of the total resistance \( R_t \) is the sum of the reciprocals of each individual resistance:\[ \frac{1}{R_t} = \frac{1}{R_1} + \frac{1}{R_2} \]Substitute the given values:\[ \frac{1}{R_t} = \frac{1}{4.00} + \frac{1}{12.0} \]This simplifies to \[ \frac{1}{R_t} = \frac{3}{12} + \frac{1}{12} = \frac{4}{12} = \frac{1}{3} \]. Therefore, the total resistance \( R_t = \frac{3}{1} = 3.00 \Omega \).
02

Include Internal Resistance of the Battery

The total circuit resistance is the sum of the total resistance of the parallel combination \( R_t \) and the internal resistance of the battery. Hence, the total resistance \( R_{total} \) is given by:\[ R_{total} = R_t + R_{internal} = 3.00 \Omega + 1.00 \Omega = 4.00 \Omega \].
03

Calculate Battery Current

Using Ohm's Law (\( V = IR \)), the current \( I_{battery} \) from the battery is determined by:\[ I_{battery} = \frac{V}{R_{total}} = \frac{22.0 \, V}{4.00 \, \Omega} = 5.50 \, A \].
04

Calculate Terminal Voltage of the Battery

The terminal voltage \( V_{terminal} \) accounts for the voltage drop across the internal resistance:\[ V_{terminal} = V - I_{battery} \times R_{internal} = 22.0 \, V - (5.50 \, A \times 1.00 \, \Omega) = 22.0 \, V - 5.50 \, V = 16.5 \, V \].
05

Calculate Current in 4.00-Ohm Resistor

Using Ohm's Law for the resistor (\( V = IR \)), where voltage across it is the terminal voltage, we find the current \( I_4 \):\[ I_4 = \frac{V_{terminal}}{R_4} = \frac{16.5 \, V}{4.00 \, \Omega} = 4.125 \, A \].
06

Calculate Current in 12.0-Ohm Resistor

Similarly, using Ohm's Law, the current \( I_{12} \) through the 12.0-ohm resistor is determined via:\[ I_{12} = \frac{V_{terminal}}{R_{12}} = \frac{16.5 \, V}{12.0 \, \Omega} = 1.375 \, A \].

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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 in electronics and physics to relate voltage, current, and resistance in an electrical circuit. The formula is simple but essential:
  • The voltage (\(V\)) across a resistor is the product of its current (\(I\)) and resistance (\(R\)).
  • It's expressed as \(V = IR\).
This equation forms the basis for analyzing circuits, such as determining the current flowing through a given resistor if the voltage across it and its resistance are known. For example, if you know that a resistor is subjected to a voltage of 16.5 volts and the resistance is 4.0 ohms, you can find the current by rearranging the formula to \(I = \frac{V}{R}\), which gives \(I = \frac{16.5}{4.0} = 4.125\, A\).
In our specific exercise: - Ohm's Law helps in finding the battery current, which has been calculated using the total resistance of the circuit and the battery's supplied voltage.- We used it to find the currents through the 4-ohm and 12-ohm resistors by considering the terminal voltage of the battery.
Internal Resistance
Internal resistance is the inherent resistance within a battery or power source that affects its performance. It effectively reduces the available voltage supplied to the external circuit.
Imagine internal resistance as a tiny obstacle within the battery cells, hindering some of the energy flow.
  • In the given exercise, the battery has an internal resistance of \(1.00\, \Omega\).
  • The presence of this resistance means that not all the 22 volts from the battery are available to the external circuit.
  • This results in a voltage drop within the battery, calculated by \(I_{battery} \times R_{internal}\), which affects the terminal voltage.
In our exercise, calculating terminal voltage involved accounting for this internal resistance by subtracting the voltage drop (\(5.50\, V\)) from the battery's original voltage. Understanding internal resistance is crucial for accurately predicting how a circuit will operate in real-world conditions.
Total Circuit Resistance
Total circuit resistance is the cumulative effect of all resistances present in a circuit, both internal and external. When resistors combine in a circuit, they can significantly alter the total resistance seen by the power source.
  • For resistors in parallel, as in our example, the total resistance (\(R_t\)) is found via the reciprocal sum of individual resistances: \(\frac{1}{R_t} = \frac{1}{R_1} + \frac{1}{R_2}\).
  • For the two resistors of \(4.00 \Omega\) and \(12.0 \Omega\), this calculation gives \(R_t = 3.00 \Omega\).
  • The circuit also includes the internal resistance of the battery, so the total circuit resistance (\(R_{total}\)) becomes the sum of the parallel resistance and the internal resistance: \(R_{total} = R_t + R_{internal} = 3.00 \Omega + 1.00 \Omega = 4.00 \Omega\).
In analyzing this circuit:- Calculating the total resistance helps determine the total current (\(I_{battery}\)) flowing from the battery by applying Ohm's Law with the battery's voltage.- This step is key in determining the load the battery sees and how much current it supplies across its terminals.

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Most popular questions from this chapter

What shunt resistance should be connected in parallel with an ammeter having a resistance of \(0.040 \Omega\) so that 25 percent of the total current will pass through the ammeter?

What resistance must be placed in parallel with \(12 \Omega\) to obtain a combined resistance of \(4 \Omega\) ? $$ \begin{array}{lc} \text { From } & \frac{1}{R_{\mathrm{eq}}} & =\frac{1}{R_{1}}+\frac{1}{R_{2}} \\\ \text { we have } & \frac{1}{4 \Omega} & =\frac{1}{12 \Omega}+\frac{1}{R_{2}} \\\ \text { and so } & R_{2}=6 \Omega \end{array} $$

A relay having a resistance of \(6.0 \Omega\) operates with a minimum current of \(0.030 \mathrm{~A}\). It is required that the relay operate when the current in the line reaches \(0.240 \mathrm{~A}\). What resistance should be used to shunt the relay?

A \(120-\mathrm{V}\) house circuit has the following light bulbs turned on: \(40.0 \mathrm{~W}, 60.0 \mathrm{~W}\), and \(175.0 \mathrm{~W}\). Find the equivalent resistance of these lights. House circuits are so constructed that each device is connected in parallel with the others. From \(\mathrm{P}=V I=V^{2} / R\), for the first bulb $$ R_{1}=\frac{V^{2}}{P_{1}}=\frac{(120 \mathrm{~V})^{2}}{40.0 \mathrm{~W}}=360 \Omega $$ Similarly, \(R_{2}=240 \Omega\) and \(R_{3}=192 \Omega\). Because devices in a house circuit are in parallel, $$ \frac{1}{R_{\mathrm{eq}}}=\frac{1}{360 \Omega}+\frac{1}{240 \Omega}+\frac{1}{192 \Omega} \quad \text { or } \quad R_{\mathrm{eq}}=82.3 \Omega $$ As a check, note that the total power drawn from the line is \(40.0 \mathrm{~W}+60.0 \mathrm{~W}+75.0 \mathrm{~W}=75.0 \mathrm{~W}\). Then, using \(\mathrm{P}=V^{2} / R\) $$ R_{\mathrm{eq}}=\frac{V^{2}}{\text { total power }}=\frac{(120 \mathrm{~V})^{2}}{175.0 \mathrm{~W}}=82.3 \Omega $$

Three resistors, of \(8.0 \Omega, 12 \Omega\), and \(24 \Omega\), are in parallel, and a current of \(20 \mathrm{~A}\) is drawn by the combination. Determine \((a)\) the potential difference across the combination and \((b)\) the current through each resistance.
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