Chapter 25: Problem 42
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.
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.
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.
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.
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.
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.