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A resistor of unknown resistance and a \(35-\Omega\) resistor are connected across a \(120-\mathrm{V}\) emf device in such a way that an 11 -A current flows. What is the value of the unknown resistance?

Short Answer

Expert verified
Answer: The value of the unknown resistor is approximately 16 Ω.

Step by step solution

01

Determine the type of connection

Since the total current flowing through the circuit is given, this implies that the resistors are connected in parallel. In a parallel connection, the total current is the sum of the currents flowing through each branch of the resistors.
02

Apply Ohm's Law to the known resistor

Ohm's Law states that the voltage across a resistor (\(V_R\)) is equal to the product of the current through the resistor (\(I_R\)) and the resistor's resistance (\(R\)). In mathematical form: \(V_R = I_RR\). We apply Ohm's Law to the 35-Ω resistor to find its current. Let's call this current \(I_1\). \(120 \mathrm{V} = I_1 (35 \Omega)\) Solving for \(I_1\), we get: \(I_1 = \dfrac{120 \mathrm{V}}{35 \Omega} = 3.43 \,\text{A}\)
03

Find the current through the unknown resistor

Since the total current in the circuit is 11-A, we can find the current through the unknown resistor by subtracting the current through the 35-Ω resistor from the total current. Let's call the current through the unknown resistor \(I_2\). \(I_2 = 11 \,\text{A} - 3.43 \,\text{A} = 7.57 \,\text{A}\)
04

Apply Ohm's Law again to find the unknown resistance

Now, we apply Ohm's Law to the unknown resistor, using the calculated current \(I_2\) and the given voltage of the emf device. Let's call the unknown resistance \(R_2\). \(120 \mathrm{V} = 7.57 \,\text{A}R_2\) Solving for \(R_2\), we get: \(R_2= \dfrac{120\mathrm{V}}{7.57 \,\text{A}} = 15.85 \,\Omega\)
05

Round off the answer to an appropriate number of significant figures

Given values in this exercise have two significant figures; hence the result should be rounded off to two significant figures. So, the value of the unknown resistance is approximately \(16\,\Omega\).

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Key Concepts

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

Resistor
A resistor is an essential component in electrical circuits, widely used to control the flow of electric current. It transforms electrical energy into heat, providing resistance measured in ohms (\(\Omega\)).
Resistors can be found in various applications from simple household electronics to complex industrial systems.
  • They are crucial in protecting sensitive components by ensuring the current doesn't exceed safe levels.
  • Resistors limit current to desirable levels, facilitating multiple functions within a circuit.
Resistors maintain their importance by contributing to functions such as voltage division, biasing active elements, and in circuitry for sensors and transducers.
Parallel Circuit
Parallel circuits allow multiple components to share the same voltage source, distributing current across various branches. They are characterized by having a clear path for current to flow through each resistor.
In a parallel circuit, the total voltage remains the same across each branch, and the total current is the sum of the currents flowing through each resistor.
  • This means each resistor receives the full voltage of the source.
  • Adding more resistors in parallel decreases the total resistance.
The beauty of parallel circuits lies in their ability to keep other components unaffected if one path fails, thereby maintaining operation in everyday devices like home lighting systems and appliances.
Current Flow
Current flow refers to the movement of electric charge through a conductor in a circuit. It is typically measured in amperes (A), and crucially determines how components like resistors influence the circuit performance.
Understanding current flow helps in calculating how much electricity moves through different parts of a circuit.
  • Ohm’s Law, expressed as \(V = IR\), is used to relate current, voltage, and resistance in circuits.
  • In parallel circuits, total current is the sum of currents through each branch, as witnessed in the given exercise.
The explanation of Ohm’s Law helps students grasp how resistors affect the flow of current, a fundamental concept for designing and analyzing electrical circuits.

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Most popular questions from this chapter

A copper wire has radius \(r=0.0250 \mathrm{~cm},\) is \(3.00 \mathrm{~m}\) long, has resistivity \(\rho=1.72 \cdot 10^{-8} \Omega \mathrm{m},\) and carries a current of \(0.400 \mathrm{~A}\). The wire has density of charge carriers of \(8.50 \cdot 10^{28}\) electrons \(/ \mathrm{m}^{3}\) a) What is the resistance, \(R,\) of the wire? b) What is the electric potential difference, \(\Delta V\), across the wire? c) What is the electric field, \(E\), in the wire?

In an emergency, you need to run a radio that uses \(30.0 \mathrm{~W}\) of power when attached to a \(10.0-\mathrm{V}\) power supply. The only power supply you have access to provides \(25.0 \mathrm{kV}\) but you do have a large number of \(25.0-\Omega\) resistors. If you want the power to the radio to be as close as possible to \(30.0 \mathrm{~W}\), how many resistors should you use, and how should they be connected (in series or in parallel)?

Show that for resistors connected in series, it is always the highest resistance that dissipates the most power, while for resistors connected in parallel, it is always the lowest resistance that dissipates the most power.

Ohm's Law states that the potential difference across a device is equal to a) the current flowing through the device times the resistance of the device. b) the current flowing through the device divided by the resistance of the device. c) the resistance of the device divided by the current flowing through the device. d) the current flowing through the device times the crosssectional area of the device. e) the current flowing through the device times the length of the device.

You are given two identical batteries and two pieces of wire. The red wire has a higher resistance than the black wire. You place the red wire across the terminals of one battery and the black wire across the terminals of the other battery. Which wire gets hotter?

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