/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 97 A group of \(N\) identical batte... [FREE SOLUTION] | 91Ó°ÊÓ

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A group of \(N\) identical batteries of \(\operatorname{emf} \mathscr{E}\) and internal resistance \(r\) may be connected all in series (Fig. \(27-80 a\) ) or all in parallel (Fig. \(27-80 b\) ) and then across a resistor \(R\). Show that both arrangements give the same current in \(R\) if \(R=r\).

Short Answer

Expert verified
Both series and parallel arrangements give the same current when \( R = r \).

Step by step solution

01

Calculate Total Voltage and Resistance for Series

In the series arrangement, the total electromotive force (emf) is the sum of the emfs of each battery, which is given by \( N \times \mathscr{E} \). The total internal resistance for the batteries is also summed up, resulting in \( N \times r \). Hence, the equivalent circuit can be represented by a single source of emf \( N \times \mathscr{E} \) and total internal resistance \( N \times r \).
02

Apply Ohm's Law for Series

Ohm's Law states that the current \( I \) through a resistor is \( I = \frac{V}{R_{total}} \), where \( V \) is the total voltage and \( R_{total} \) is the total resistance. For the series arrangement:\[ I_{series} = \frac{N \times \mathscr{E}}{N \times r + R} \].
03

Calculate Total Voltage and Resistance for Parallel

In the parallel arrangement, the total emf stays the same, \( \mathscr{E} \), since voltage across each parallel branch remains unchanged. The internal resistances, however, are in parallel, leading to equivalent internal resistance:\[ r_{eq} = \frac{r}{N} \]. Thus, the equivalent circuit is a source with emf \( \mathscr{E} \) and resistance \( r_{eq} \).
04

Apply Ohm's Law for Parallel

Using Ohm's Law for the parallel circuit setup, we find the current \( I \) through the external resistor \( R \):\[ I_{parallel} = \frac{\mathscr{E}}{\frac{r}{N} + R} \].
05

Compare Currents for Series and Parallel

Set \( R = r \) and compare the two current expressions:- For series: \[ I_{series} = \frac{N \times \mathscr{E}}{N \times r + r} = \frac{N \times \mathscr{E}}{(N+1) \times r} \]- For parallel: \[ I_{parallel} = \frac{\mathscr{E}}{\frac{r}{N} + r} = \frac{N \times \mathscr{E}}{r+N \times r} = \frac{N \times \mathscr{E}}{(N+1) \times r} \]Both currents are equal demonstrating that both arrangements yield the same current when \( R = r \).

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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 in electrical circuits that relates the current flowing through a conductor to the voltage across it and its resistance. It is expressed with the formula: \[ I = \frac{V}{R} \] where:
  • \( I \) is the current in amperes (A),
  • \( V \) is the voltage in volts (V), and
  • \( R \) is the resistance in ohms (Ω).
This law helps us to understand how the voltage and resistance in a circuit influence the current.
In series circuits, the total voltage is the sum of all individual voltages, while in parallel circuits, the voltage across each branch is the same.
This indicates how Ohm's Law can be used differently depending on the circuit configuration. Furthermore, understanding Ohm's Law is critical when calculating the current produced in any circuit arrangement. It's especially useful when these circuits involve components with internal resistance, like batteries.
Series Circuits
In series circuits, components are connected end to end, forming a single path for the current to flow.
This means that the same current travels through each component. The total voltage across a series circuit is the sum of the voltages across each component:\[ V_{total} = V_1 + V_2 + V_3 + \, ... \, + V_n \] where each \( V_n \) represents the voltage across individual components.

The total resistance in a series circuit is also the sum of all individual resistances:\[ R_{total} = R_1 + R_2 + R_3 + \, ... \, + R_n \]This implies that as more components are added in series, the total resistance increases, affecting the overall current according to Ohm's Law.
Therefore, this configuration is useful when uniform current through multiple components is essential.
Understanding series circuits is crucial when assessing battery setups, as the total emf and resistance both increase with each additional battery.
Parallel Circuits
Parallel circuits provide multiple paths for current to flow.
Unlike in series circuits, voltage across each component in parallel is the same as the source voltage.The total equivalent resistance in a parallel circuit can be found using the formula:\[ \frac{1}{R_{total}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \, ... \, + \frac{1}{R_n} \]This results in a lower total resistance compared to the individual resistances, making parallel circuits effective for reducing total resistance in a network.

In a parallel arrangement of batteries, identical batteries maintain the same emf across all branches, but the internal resistances add up inversely.
With a lower equivalent resistance, more current can flow through the circuit compared to a series configuration.
This principle is applied when a stable voltage is needed across all parts of a circuit while allowing for a larger total current flow.
Internal Resistance
Internal resistance is the resistance within a battery or power source that impedes the flow of charge.
It is a critical factor to consider as it affects the output of the battery when in use. Every battery has an internal resistance denoted by \( r \), and this becomes significant especially when batteries are used in circuits.When batteries are connected in series, their internal resistances add up:\[ R_{internal\, series} = r_1 + r_2 + r_3 + \, ... \, + r_n \] In contrast, when connected in parallel, the total internal resistance decreases: \[ \frac{1}{R_{internal\, parallel}} = \frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} + \, ... \, + \frac{1}{r_n} \]Thus, minimizing the impact of internal resistance is essential for efficient power delivery.
For this reason, knowing how to calculate and account for internal resistance can significantly affect the performance of electrical devices, as well as the total current flowing through circuits.

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Most popular questions from this chapter

A simple ohmmeter is made by connecting a \(1.50 \mathrm{~V}\) flashlight battery in series with a resistance \(R\) and an ammeter that reads from 0 to \(1.00 \mathrm{~mA}\), as shown in Fig. \(27-59 .\) Resistance \(R\) is adjusted so that when the clip leads are shorted together, the meter deflects to its full- scale value of \(1.00 \mathrm{~mA}\). What external resistance across the leads results in a deflection of (a) \(10.0 \%,(\mathrm{~b})\) \(50.0 \%\), and (c) \(90.0 \%\) of full scale? (d) If the ammeter has a resistance of \(20.0 \Omega\) and the internal resistance of the battery is negligible, what is the value of \(R ?\)

Two identical batteries of emf \(\mathscr{E}=12.0\) \(V\) and internal resistance \(r=0.200 \Omega\) are to be connected to an external resistance \(R\), either in parallel (Fig. \(27-50\) ) or in series (Fig. 27-51). If \(R=2.00 r\), what is the current \(i\) in the external resistance in the (a) parallel and (b) series arrangements? (c) For which arrangement is \(i\) greater? If \(R=r / 2.00\), what is \(i\) in the external resistance in the (d) parallel and (e) series arrangements? (f) For which arrangement is \(i\) greater now?

A car battery with a \(12 \mathrm{~V}\) emf and an internal resistance of \(0.040\) \(\Omega\) is being charged with a current of 50 A. What are (a) the potential difference \(V\) across the terminals, (b) the rate \(P_{r}\) of energy dissipation inside the battery, and (c) the rate \(P_{\text {emr }}\) of energy conversion to chemical form? When the battery is used to supply 50 A to the starter motor, what are (d) \(V\) and (e) \(P_{r}\) ?

A certain car battery with a \(12.0 \mathrm{~V}\) emf has an initial charge of \(120 \mathrm{~A} \cdot \mathrm{h}\). Assuming that the potential across the terminals stays constant until the battery is completely discharged, for how many hours can it deliver energy at the rate of \(100 \mathrm{~W}\) ?

What multiple of the time constant \(\tau\) gives the time taken by an initially uncharged capacitor in an \(R C\) series circuit to be charged to \(99.0 \%\) of its final charge?

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