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Magazine Chapter 17: Problem 11
(a) What current does a \(150-\Omega\) resistance draw on a \(1 \overline{0}-\mathrm{V}\) battery? (b) What voltage battery would produce three times the current in (a)? (c) What current would a \(75-\Omega\) resistor draw on the \(1 \overline{0}-\mathrm{V}\) battery?
Short Answer
Expert verified
(a) 0.733 A; (b) 329.85 V; (c) 1.467 A.
Step by step solution
01
Identify Given Values for Part (a)
We are given a resistance \(R_1 = 150\, \Omega\) and a voltage \(V = 110\, \text{V}\). We need to find the current \(I_1\) flowing through this resistor.
02
Use Ohm's Law for Part (a)
Ohm's Law relates current \(I\), voltage \(V\), and resistance \(R\) with the equation \(I = \frac{V}{R}\). Substitute the given values: \(I_1 = \frac{110}{150}\).
03
Calculate the Current for Part (a)
Perform the division: \(I_1 = \frac{110}{150} = 0.733\, \text{A}\). Thus, the current is approximately \(0.733\, \text{A}\).
04
Determine the Required Current for Part (b)
In part (b), we want the current to be three times that of part (a). So, the required current \(I_2 = 3 \times 0.733 = 2.199\, \text{A}\).
05
Use Ohm's Law to Find Battery Voltage for Part (b)
Using Ohm's Law \(V = I_2 \times R_1\), substitute the values \(V = 2.199 \times 150\).
06
Calculate the Required Voltage for Part (b)
Perform the multiplication: \(V = 329.85\, \text{V}\). Thus, a voltage of \(329.85\, \text{V}\) is needed to produce the desired current.
07
Identify Given Values for Part (c)
We are given a resistance \(R_2 = 75\, \Omega\) and voltage \(V = 110\, \text{V}\). We need to find the current \(I_3\) flowing through this resistor.
08
Use Ohm's Law for Part (c)
Using Ohm's Law \(I = \frac{V}{R}\), substitute \(V = 110\) and \(R_2 = 75\): \(I_3 = \frac{110}{75}\).
09
Calculate the Current for Part (c)
Perform the division: \(I_3 = \frac{110}{75} = 1.467\, \text{A}\). Therefore, the current is approximately \(1.467\, \text{A}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Understanding Electrical Resistance
Electrical resistance is a fundamental concept in circuits that measures how much a material opposes the flow of electric current. It is like a bumpy road slowing down a car. Resistance determines how much current flows for a given voltage. The unit of resistance is the ohm (\( \Omega \)). More resistance means less current flow for the same voltage.
- Materials like copper have low resistance, making them good conductors of electricity.
- Insulators like rubber have high resistance, preventing current from flowing easily.
- Type of material
- Temperature
- Length and thickness of the material
Calculating Current Flow
Current calculation is an essential part of understanding how electricity flows through a circuit. Using Ohm's Law, which states that current (\( I \)) is equal to voltage (\( V \)) divided by resistance (\( R \)), we can easily find out how much current flows in a circuit.
The basic formula is: \[I = \frac{V}{R}\]In the exercise:
The basic formula is: \[I = \frac{V}{R}\]In the exercise:
- For a 150-ohm resistor with a 110-volt battery, the current is calculated as \( 0.733 \) amperes.
- For a 75-ohm resistor with the same battery, the current is \( 1.467 \) amperes. This is more than the current through the 150-ohm resistor, as expected with the lower resistance.
- Part (b) challenges us by asking what voltage is needed to get three times the current of part (a), leading to a calculation of a 329.85-volt battery.
Exploring Battery Voltage
Battery voltage is the driving force pushing electric current through a circuit. When you think of voltage, imagine it as the pressure pushing water through a pipe. The higher the voltage, the greater the force available to push electrons through a resistive path.
In the exercise, we began with a standard battery voltage of 110 volts and changed it to explore different scenarios. We wanted to see how it affected current flow in resistors.
In the exercise, we began with a standard battery voltage of 110 volts and changed it to explore different scenarios. We wanted to see how it affected current flow in resistors.
- With resistance kept constant, increasing voltage increases current. This is seen in part (b) where a higher voltage was needed to triple the current through a 150-ohm resistor.
- This concept is crucial when designing circuits to ensure components withstand desired operation ranges without damage.
