9. an electric motor has an apparent resistance of 15 ω. if 8 a of current are flowing through the motor, what is the connected voltage?
10. a 240 - v air conditioning compressor has an apparent resistance of 8 ω. how much current will flow in the circuit?
11. how much power is being used by the compressor in question 8?
12. a 5 - kw electric heating unit is connected to a 240 - v line. what is the current flow in the circuit?
13. if the voltage in question 12 is reduced to 120 v, how much current would be needed to reduce the same amount of power?
14. is it less expensive to operate the electric heating unit in question 12 on 240 v or 120 v?
15. name three types of fixed resistors.
16. what is the advantage of a metal film resistor over a carbon resistor?
17. what is the advantage of a wire wound resistor?
18. how should tubular wire wound resistors be mounted and why?
19. a half - watt 2,000 - ω resistor has a current flow of.01 amps through it. is this resistor operating within its power rating?
20. a one - watt 350 - ω resistor is connected to 24 v. is this resistor operating within its power rating?
21. a resistor has color bands of orange, blue, yellow, and gold. what is the resistance and tolerance of this resistor?
22. a 10,000 - ω resistor has a tolerance of 5%. what are the minimum and maximum ratings of this resistor?
23. is 51,000 ω a standard value for a 5% resistor?
24. what is a potentiometer?

Question

Accepted Answer

Let's solve question 9 first (as an example, since you can choose any sub-question).

### Question 9:
# Explanation:
## Step1: Recall Ohm's Law
Ohm's Law is given by $ V = I \times R $, where $ V $ is voltage, $ I $ is current, and $ R $ is resistance.
## Step2: Identify given values
We are given $ R = 15\ \Omega $ and $ I = 8\ A $.
## Step3: Substitute values into the formula
Substitute $ I = 8\ A $ and $ R = 15\ \Omega $ into $ V = I \times R $. So, $ V = 8\ A \times 15\ \Omega $.
## Step4: Calculate the voltage
$ 8 \times 15 = 120 $, so $ V = 120\ V $.
# Answer:
The connected voltage is $ 120\ V $.

### Question 10:
# Explanation:
## Step1: Recall Ohm's Law
Ohm's Law is $ I=\frac{V}{R} $, where $ V $ is voltage, $ I $ is current, and $ R $ is resistance.
## Step2: Identify given values
Given $ V = 240\ V $ and $ R = 8\ \Omega $.
## Step3: Substitute values into the formula
Substitute $ V = 240\ V $ and $ R = 8\ \Omega $ into $ I=\frac{V}{R} $. So, $ I=\frac{240\ V}{8\ \Omega} $.
## Step4: Calculate the current
$ \frac{240}{8}=30 $, so $ I = 30\ A $.
# Answer:
The current flowing in the circuit is $ 30\ A $.

### Question 11: (Assuming question 8 is similar, but since it's not shown, let's assume the compressor in question 10. So using $ V = 240\ V $, $ I = 30\ A $ from question 10)
# Explanation:
## Step1: Recall Power formula (for electrical circuits)
Power $ P = V \times I $, where $ V $ is voltage and $ I $ is current.
## Step2: Identify given values
From question 10, $ V = 240\ V $ and $ I = 30\ A $.
## Step3: Substitute values into the formula
Substitute $ V = 240\ V $ and $ I = 30\ A $ into $ P = V \times I $. So, $ P = 240\ V \times 30\ A $.
## Step4: Calculate the power
$ 240\times30 = 7200 $, so $ P = 7200\ W = 7.2\ kW $.
# Answer:
The power used by the compressor is $ 7200\ W $ (or $ 7.2\ kW $).

### Question 12:
# Explanation:
## Step1: Recall Power formula and rearrange for current
Power $ P = V \times I $, so we can rearrange it to $ I=\frac{P}{V} $. Given $ P = 5\ kW = 5000\ W $ and $ V = 240\ V $.
## Step2: Substitute values into the formula
Substitute $ P = 5000\ W $ and $ V = 240\ V $ into $ I=\frac{P}{V} $. So, $ I=\frac{5000\ W}{240\ V} $.
## Step3: Calculate the current
$ \frac{5000}{240}\approx20.83\ A $ (rounded to two decimal places).
# Answer:
The current flow in the circuit is approximately $ 20.83\ A $.

### Question 13:
# Explanation:
## Step1: Recall Power formula and rearrange for current
Power $ P = V \times I $, so $ I=\frac{P}{V} $. The power $ P $ is still $ 5\ kW = 5000\ W $ (same amount of power), and the new voltage $ V = 120\ V $.
## Step2: Substitute values into the formula
Substitute $ P = 5000\ W $ and $ V = 120\ V $ into $ I=\frac{P}{V} $. So, $ I=\frac{5000\ W}{120\ V} $.
## Step3: Calculate the current
$ \frac{5000}{120}\approx41.67\ A $ (rounded to two decimal places).
# Answer:
The current needed is approximately $ 41.67\ A $.

### Question 14:
# Explanation:
## Step1: Recall the cost of electricity is related to power and time (but since power is the same, we can think about current and resistance losses, but more simply, for a resistive load, power $ P=\frac{V^{2}}{R} $, but also, the power loss in wires (if we consider the circuit from the source to the load) is $ I^{2}R_{wire} $. However, for the heating unit (a resistive load), $ R=\frac{V^{2}}{P} $. First, find the resistance of the heating unit from question 12. From $ P = 5000\ W $ and $ V = 240\ V $, $ R=\frac{V^{2}}{P}=\frac{(240)^{2}}{5000}=\frac{57600}{5000}=11.52\ \Omega $.
## Step2: Calculate power loss in wires (assuming the wire resistance is constant, say $ R_{wire} $ is some value, but actually, the main idea is that for the same power, $ I=\frac{P}{V} $, so lower voltage means higher current. Higher current means more power loss in the wires (since $ P_{loss}=I^{2}R_{wire} $). So, operating at 240 V has lower current ($ \approx20.83\ A $) than at 120 V ($ \approx41.67\ A $). So, less power loss in the circuit when operating at 240 V, making it less expensive (since we pay for the power used by the load plus the power lost in the wires; the load power is the same, but wire loss is less at higher voltage for the same load power).
## Step3: Conclusion
Since the current is lower at 240 V, the power loss in the connecting wires (if any) will be less (because $ P_{loss}=I^{2}R $), so it is less expensive to operate the electric heating unit at 240 V.
# Answer:
It is less expensive to operate the electric heating unit on 240 V.

### Question 15:
# Brief Explanations:
Fixed resistors are resistors with a fixed resistance value. Three common types are: 1. Carbon - composition resistors: Made of a mixture of carbon and a binding material. 2. Metal - film resistors: Have a thin metal film (like nickel - chromium) on a ceramic substrate. 3. Carbon - film resistors: Have a carbon film deposited on a ceramic substrate.
# Answer:
Three types of fixed resistors are carbon - composition resistors, metal - film resistors, and carbon - film resistors.

### Question 16:
# Brief Explanations:
Metal - film resistors have several advantages over carbon resistors. Metal - film resistors have better stability (their resistance value changes less with temperature and over time), higher precision (more accurate resistance values), and lower noise (they produce less electrical noise) compared to carbon resistors.
# Answer:
Metal - film resistors have better stability, higher precision, and lower noise than carbon resistors.

### Question 17:
# Brief Explanations:
Wire - wound resistors have the advantage of being able to handle high power dissipation. They can be made with very precise resistance values, especially for low - resistance and high - power applications. Also, they have good stability over a wide range of temperatures.
# Answer:
Wire - wound resistors can handle high power, have precise resistance values, and good temperature stability.

### Question 18:
# Brief Explanations:
Tubular wire - wound resistors should be mounted vertically (on their ends) or with proper spacing to allow air to circulate around them. This is because they dissipate a significant amount of heat, and mounting them in a way that allows air to flow helps in dissipating the heat effectively, preventing overheating and damage to the resistor or nearby components.
# Answer:
Tubular wire - wound resistors should be mounted vertically or with spacing for air circulation to dissipate heat.

### Question 19:
# Explanation:
## Step1: Recall Power formula for a resistor
The power dissipated by a resistor is given by $ P = I^{2}R $, where $ I $ is the current and $ R $ is the resistance.
## Step2: Identify given values
Given $ I = 0.01\ A $, $ R = 2000\ \Omega $, and the resistor's power rating is $ 0.5\ W $ (half - watt).
## Step3: Substitute values into the formula
Substitute $ I = 0.01\ A $ and $ R = 2000\ \Omega $ into $ P = I^{2}R $. So, $ P=(0.01\ A)^{2}\times2000\ \Omega $.
## Step4: Calculate the power
First, $ (0.01)^{2}=0.0001 $, then $ 0.0001\times2000 = 0.2\ W $.
## Step5: Compare with the power rating
The calculated power $ 0.2\ W $ is less than the power rating of $ 0.5\ W $.
# Answer:
Yes, the resistor is operating within its power rating.

### Question 20:
# Explanation:
## Step1: Recall Power formula for a resistor
The power dissipated by a resistor is $ P=\frac{V^{2}}{R} $, where $ V $ is the voltage and $ R $ is the resistance.
## Step2: Identify given values
Given $ V = 24\ V $, $ R = 350\ \Omega $, and the resistor's power rating is $ 1\ W $.
## Step3: Substitute values into the formula
Substitute $ V = 24\ V $ and $ R = 350\ \Omega $ into $ P=\frac{V^{2}}{R} $. So, $ P=\frac{(24\ V)^{2}}{350\ \Omega} $.
## Step4: Calculate the power
First, $ (24)^{2}=576 $, then $ \frac{576}{350}\approx1.646\ W $.
## Step5: Compare with the power rating
The calculated power $ \approx1.646\ W $ is greater than the power rating of $ 1\ W $.
# Answer:
No, the resistor is not operating within its power rating.

### Question 21:
# Explanation:
## Step1: Recall resistor color - code rules
The first two bands are significant figures, the third band is the multiplier (number of zeros), and the fourth band is the tolerance.
- Orange represents 3, blue represents 6, yellow represents $ 10^{4} $, and gold represents a tolerance of $ \pm5\% $.
## Step2: Calculate the resistance
Significant figures: $ 3$ (orange) and $ 6$ (blue). Multiplier: $ 10^{4}$ (yellow). So, resistance $ R = 36\times10^{4}\ \Omega=360000\ \Omega = 360\ k\Omega $. Tolerance: $ \pm5\% $ (from gold band).
# Answer:
The resistance of the resistor is $ 360000\ \Omega $ (or $ 360\ k\Omega $) with a tolerance of $ \pm5\% $.

### Question 22:
# Explanation:
## Step1: Recall tolerance formula
The minimum resistance $ R_{min}=R - (R\times\text{tolerance}) $ and the maximum resistance $ R_{max}=R+(R\times\text{tolerance}) $, where $ R = 10000\ \Omega $ and tolerance $ = 5\%=0.05 $.
## Step2: Calculate minimum resistance
$ R_{min}=10000\ \Omega-(10000\ \Omega\times0.05)=10000 - 500 = 9500\ \Omega $.
## Step3: Calculate maximum resistance
$ R_{max}=10000\ \Omega+(10000\ \Omega\times0.05)=10000 + 500 = 10500\ \Omega $.
# Answer:
The minimum rating of the resistor is $ 9500\ \Omega $ and the maximum rating is $ 10500\ \Omega $.

### Question 23:
# Brief Explanations:
Standard resistor values for $ 5\% $ tolerance (E6 series or E12 series, but for $ 5\% $ tolerance, the E24 series has more values, but common standard values around 51000 Ω: Wait, 51000 Ω is $ 51\ k\Omega $. In the E24 series (5% tolerance), the standard values include 51 kΩ. So, yes, 51000 Ω is a standard value for a 5% resistor.
# Answer:
Yes, $ 51000\ \Omega $ (or $ 51\ k\Omega $) is a standard value for a 5% resistor.

### Question 24:
# Brief Explanations:
A potentiometer is a three - terminal resistor with a sliding or rotating contact that forms an adjustable voltage divider. It can be used to adjust the resistance in a circuit, thereby controlling the voltage or current in the circuit. For example, it is used in volume controls of audio devices, to adjust the brightness of lights in some applications, etc.
# Answer:
A potentiometer is a three - terminal adjustable resistor that functions as a voltage divider to control voltage or current in a circuit.

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Question 9:

Step1: Recall Ohm's Law

Step2: Identify given values

Step3: Substitute values into the formula

Step4: Calculate the voltage

Step1: Recall Ohm's Law

Step2: Identify given values

Step3: Substitute values into the formula

Step4: Calculate the current

Step1: Recall Power formula (for electrical circuits)

Step2: Identify given values

Step3: Substitute values into the formula

Step4: Calculate the power

Answer:

Question 10: