Problem 11 - Applied Physics

Question

(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?

Step-by-Step Solution

Answer

(a) 0.733 A; (b) 329.85 V; (c) 1.467 A.

Step 1: 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.

Step 2: 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: \(I_1 = \frac{110}{150}\).

Step 3: Calculate the Current for Part (a)

Perform the division: \(I_1 = \frac{110}{150} = 0.733\, \text{A}\). Thus, the current is approximately \(0.733\, \text{A}\).

Step 4: 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 \(I_2 = 3 \times 0.733 = 2.199\, \text{A}\).

Step 5: Use Ohm's Law to Find Battery Voltage for Part (b)

Using Ohm's Law \(V = I_2 \times R_1\), substitute the values \(V = 2.199 \times 150\).

Step 6: 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.

Step 7: 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.

Step 8: Use Ohm's Law for Part (c)

Using Ohm's Law \(I = \frac{V}{R}\), substitute \(V = 110\) and \(R_2 = 75\): \(I_3 = \frac{110}{75}\).

Step 9: Calculate the Current for Part (c)

Perform the division: \(I_3 = \frac{110}{75} = 1.467\, \text{A}\). Therefore, the current is approximately \(1.467\, \text{A}\).

Key Concepts

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.

Resistance in a circuit can be influenced by:

• Type of material
• Temperature
• Length and thickness of the material

In the exercise, we calculated current using two different resistances: 150 ohms and 75 ohms. The lower resistance (75 ohms) allowed more current to flow compared to the higher resistance (150 ohms) when connected to the same voltage.

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:

• 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.

This demonstrates the direct relationship between voltage, current, and resistance in electrical circuits.

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.

• 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.

By understanding battery voltage and how it influences current and resistance, you gain insight into optimizing circuit design for a variety of electronic applications.