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Chapter 27: Problem 63

The current in a single-loop circuit with one resistance \(R\) is \(5.0 \mathrm{~A}\). When an additional resistance of \(2.0 \Omega\) is inserted in series with \(R\), the current drops to \(3.0 \mathrm{~A}\). What is \(R\) ?

Short Answer

Expert verified
The resistance \(R\) is \(3 \Omega\).

Step by step solution

01

Determine the initial condition

When there is only resistance \(R\) in the circuit, the current is \(5.0 \mathrm{~A}\). Applying Ohm鈥檚 Law, which states \(V = IR\), we assume a voltage \(V\) which is constant throughout: \(V = 5R\).
02

Establish the condition with additional resistance

When an additional resistance of \(2.0 \Omega\) is added, the total resistance becomes \(R + 2.0 \Omega\), and the current drops to \(3.0 \mathrm{~A}\). Again using Ohm's Law: \(V = 3(R + 2)\).
03

Equate the expressions for voltage

Since the voltage \(V\) remains the same in both cases, we equate the two expressions: \(5R = 3(R + 2)\).
04

Solve the equation for \(R\)

Expand the equation: \(5R = 3R + 6\). Rearranging gives: \(5R - 3R = 6\). Simplify to find: \(2R = 6\). Therefore, \(R = \frac{6}{2} = 3 \Omega\).

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Key Concepts

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

Resistance in Series
Whenever we discuss resistance in series, think of it as linking several resistors end-to-end, like a chain. In a series configuration, all resistors are aligned in one single pathway for the electric current. This results in a cumulative resistance.
Here's why this is important:
  • The total or equivalent resistance, denoted as \( R_{total} \), is simply the sum of all individual resistances in the series. That means, for resistors \( R_1 \), \( R_2 \), and \( R_3 \) connected in series, the total resistance is \( R_{total} = R_1 + R_2 + R_3 \).
  • The same current flows through each resistor because the pathway for current is continuous and unbroken.
When new resistance is added to a circuit already in series, the total resistance will increase. As seen in the original exercise, adding more resistance reduces the overall current due to the increased opposition to the flow of current. This is a key reason why the initial current drops when a new resistor is added.
Current in Circuits
Current refers to the flow of electric charge through a circuit and is measured in amperes (A). Think of current like water flowing through a pipe; the wider the pipe, the more water can flow through per unit time. In the case of electric circuits, the 'width' is determined by the resistance.
  • Ohm's Law, which is a crucial equation in understanding current flow, is given by \( V = IR \), where \( V \) is voltage, \( I \) is current, and \( R \) is resistance.
  • In simple terms, if you know the voltage across a circuit and its resistance, you can calculate the current.
When resistance increases, like when resistors are added in series, the current decreases if the voltage remains constant. This is because, given a fixed voltage, the flow of current is inversely affected by total resistance.
Voltage Drop
The voltage drop is an important concept in circuit analysis, especially when dealing with resistors in series. Voltage drop refers to the reduction in voltage as electrical energy is used up by resistors in a circuit.
Here's how it works:
  • As current flows through each resistor, it encounters resistance which causes a loss of electrical energy. This loss appears as a voltage drop across the resistor.
  • In a series circuit, the sum of the individual voltage drops across each resistor equals the total voltage supplied by the source.
In the exercise, as an additional resistor is added, the total resistance increases, causing the overall current in the circuit to drop. This change affects the voltage drop as the resistors share the same total voltage but now have a higher total resistance. Thus, the value of each individual resistor affects how much voltage drop each one will experience.

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Most popular questions from this chapter

A capacitor with initial charge \(q_{0}\) is discharged through a resistor. What multiple of the time constant \(\tau\) gives the time the capacitor takes to lose (a) the first \(25 \%\) of its charge and (b) \(50 \%\) of its charge?

Two identical batteries of emf \(8=10.0 \mathrm{~V} \quad\) Problems and internal resistance \(r=0.200 \Omega\) are to be 5 and 6 . connected to an external resistance \(R\), either in parallel (Fig. 27-20) or in series (Fig. 27-21). If \(R=2.00 \mathrm{r}\), 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 arrangement and (e) series arrangement? (f) For which arrangement is \(i\) greater now?

A standard flashlight battery can deliver about \(2.0 \mathrm{~W} \cdot \mathrm{h}\) of energy before it runs down. (a) If a battery costs US\$ \(0.85\), what is the cost of operating a \(100 \mathrm{~W}\) lamp for \(8.0 \mathrm{~h}\) using batteries? (b) What is the cost if energy is provided at the rate of US\$0.06 per kilowatt-hour?

A \(1.0 \mu \mathrm{F}\) capacitor with an initial stored energy of \(0.60 \mathrm{~J}\) is discharged through a \(1.0 \mathrm{M} \Omega\) resistor. (a) What is the initial charge on the capacitor? (b) What is the current through the resistor when the discharge starts? Find an expression that gives, as a function of time \(t\), (c) the potential difference \(V_{c}\) across the capacitor, (d) the potential difference \(V_{R}\) across the resistor, and (e) the rate at which thermal energy is produced in the resistor.

A voltmeter of resistance \(R_{\mathrm{V}}=300 \Omega\) and an ammeter of resistance \(R_{A}=3.00 \Omega\) are being used to measure a resistance \(R\) in a circuit that also contains a resistance \(R_{0}=100 \Omega\) and an ideal battery with an emf of \(\mathscr{\&}=18.0 \mathrm{~V}\). Resistance \(R\) is given by \(R=V / i\), where \(V\) is the potential across \(R\) and \(i\) is the ammeter read- ing. The voltmeter reading is \(V^{\prime}\), which is \(V\) plus the potential difference across the ammeter. Thus, the ratio of the two meter readings is not \(R\) but only an apparent resistance \(R^{\prime}=V^{\prime} / i\). If \(R=85.0 \Omega\), what are (a) the ammeter reading, (b) the voltmeter reading, and (c) \(R^{\prime} ?\) (d) If \(R_{A}\) is decreased, does the difference between \(R^{\prime}\) and \(R\) increase, decrease, or remain the same?
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