Welcome to Westonci.ca, the Q&A platform where your questions are met with detailed answers from experienced experts. Explore a wealth of knowledge from professionals across various disciplines on our comprehensive Q&A platform. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform.
Sagot :
To solve this problem, we need to check which circuits deliver more than [tex]\(4.0 \, \text{C}\)[/tex] of charge in [tex]\(8.0 \, \text{s}\)[/tex].
We can use the following steps:
1. Calculate the current [tex]\(I\)[/tex] for each circuit using Ohm's Law:
[tex]\[ I = \frac{V}{R} \][/tex]
where [tex]\(I\)[/tex] is the current, [tex]\(V\)[/tex] is the voltage, and [tex]\(R\)[/tex] is the resistance.
2. Calculate the charge [tex]\(Q\)[/tex] delivered in [tex]\(8.0 \, \text{s}\)[/tex] using the formula:
[tex]\[ Q = I \cdot t \][/tex]
where [tex]\(Q\)[/tex] is the charge, [tex]\(I\)[/tex] is the current, and [tex]\(t\)[/tex] is the time.
3. Compare the calculated charge to the threshold [tex]\(4.0 \, \text{C}\)[/tex] to determine if it exceeds this value.
Now let's go through the circuits one by one:
### Circuit W
- Voltage [tex]\(V = 18 \, \text{V}\)[/tex]
- Resistance [tex]\(R = 38 \, \Omega\)[/tex]
- Time [tex]\(t = 8.0 \, \text{s}\)[/tex]
Calculate the current:
[tex]\[ I_W = \frac{18}{38} \approx 0.474 \, \text{A} \][/tex]
Calculate the charge:
[tex]\[ Q_W = 0.474 \cdot 8.0 \approx 3.792 \, \text{C} \][/tex]
Since [tex]\(3.792 \, \text{C}\)[/tex] is less than [tex]\(4.0 \, \text{C}\)[/tex], Circuit [tex]\(W\)[/tex] does not need to be reported.
### Circuit X
- Voltage [tex]\(V = 24 \, \text{V}\)[/tex]
- Resistance [tex]\(R = 34 \, \Omega\)[/tex]
- Time [tex]\(t = 8.0 \, \text{s}\)[/tex]
Calculate the current:
[tex]\[ I_X = \frac{24}{34} \approx 0.706 \, \text{A} \][/tex]
Calculate the charge:
[tex]\[ Q_X = 0.706 \cdot 8.0 \approx 5.648 \, \text{C} \][/tex]
Since [tex]\(5.648 \, \text{C}\)[/tex] is greater than [tex]\(4.0 \, \text{C}\)[/tex], Circuit [tex]\(X\)[/tex] needs to be reported.
### Circuit Y
- Voltage [tex]\(V = 34 \, \text{V}\)[/tex]
- Resistance [tex]\(R = 70 \, \Omega\)[/tex]
- Time [tex]\(t = 8.0 \, \text{s}\)[/tex]
Calculate the current:
[tex]\[ I_Y = \frac{34}{70} \approx 0.486 \, \text{A} \][/tex]
Calculate the charge:
[tex]\[ Q_Y = 0.486 \cdot 8.0 \approx 3.888 \, \text{C} \][/tex]
Since [tex]\(3.888 \, \text{C}\)[/tex] is less than [tex]\(4.0 \, \text{C}\)[/tex], Circuit [tex]\(Y\)[/tex] does not need to be reported.
### Circuit Z
- Voltage [tex]\(V = 12 \, \text{V}\)[/tex]
- Resistance [tex]\(R = 18 \, \Omega\)[/tex]
- Time [tex]\(t = 8.0 \, \text{s}\)[/tex]
Calculate the current:
[tex]\[ I_Z = \frac{12}{18} \approx 0.667 \, \text{A} \][/tex]
Calculate the charge:
[tex]\[ Q_Z = 0.667 \cdot 8.0 \approx 5.336 \, \text{C} \][/tex]
Since [tex]\(5.336 \, \text{C}\)[/tex] is greater than [tex]\(4.0 \, \text{C}\)[/tex], Circuit [tex]\(Z\)[/tex] needs to be reported.
### Conclusion
The circuits that need to be reported are [tex]\(X\)[/tex] and [tex]\(Z\)[/tex]. Therefore, the correct answer is:
[tex]\[ \boxed{X \text{ and } Z} \][/tex]
We can use the following steps:
1. Calculate the current [tex]\(I\)[/tex] for each circuit using Ohm's Law:
[tex]\[ I = \frac{V}{R} \][/tex]
where [tex]\(I\)[/tex] is the current, [tex]\(V\)[/tex] is the voltage, and [tex]\(R\)[/tex] is the resistance.
2. Calculate the charge [tex]\(Q\)[/tex] delivered in [tex]\(8.0 \, \text{s}\)[/tex] using the formula:
[tex]\[ Q = I \cdot t \][/tex]
where [tex]\(Q\)[/tex] is the charge, [tex]\(I\)[/tex] is the current, and [tex]\(t\)[/tex] is the time.
3. Compare the calculated charge to the threshold [tex]\(4.0 \, \text{C}\)[/tex] to determine if it exceeds this value.
Now let's go through the circuits one by one:
### Circuit W
- Voltage [tex]\(V = 18 \, \text{V}\)[/tex]
- Resistance [tex]\(R = 38 \, \Omega\)[/tex]
- Time [tex]\(t = 8.0 \, \text{s}\)[/tex]
Calculate the current:
[tex]\[ I_W = \frac{18}{38} \approx 0.474 \, \text{A} \][/tex]
Calculate the charge:
[tex]\[ Q_W = 0.474 \cdot 8.0 \approx 3.792 \, \text{C} \][/tex]
Since [tex]\(3.792 \, \text{C}\)[/tex] is less than [tex]\(4.0 \, \text{C}\)[/tex], Circuit [tex]\(W\)[/tex] does not need to be reported.
### Circuit X
- Voltage [tex]\(V = 24 \, \text{V}\)[/tex]
- Resistance [tex]\(R = 34 \, \Omega\)[/tex]
- Time [tex]\(t = 8.0 \, \text{s}\)[/tex]
Calculate the current:
[tex]\[ I_X = \frac{24}{34} \approx 0.706 \, \text{A} \][/tex]
Calculate the charge:
[tex]\[ Q_X = 0.706 \cdot 8.0 \approx 5.648 \, \text{C} \][/tex]
Since [tex]\(5.648 \, \text{C}\)[/tex] is greater than [tex]\(4.0 \, \text{C}\)[/tex], Circuit [tex]\(X\)[/tex] needs to be reported.
### Circuit Y
- Voltage [tex]\(V = 34 \, \text{V}\)[/tex]
- Resistance [tex]\(R = 70 \, \Omega\)[/tex]
- Time [tex]\(t = 8.0 \, \text{s}\)[/tex]
Calculate the current:
[tex]\[ I_Y = \frac{34}{70} \approx 0.486 \, \text{A} \][/tex]
Calculate the charge:
[tex]\[ Q_Y = 0.486 \cdot 8.0 \approx 3.888 \, \text{C} \][/tex]
Since [tex]\(3.888 \, \text{C}\)[/tex] is less than [tex]\(4.0 \, \text{C}\)[/tex], Circuit [tex]\(Y\)[/tex] does not need to be reported.
### Circuit Z
- Voltage [tex]\(V = 12 \, \text{V}\)[/tex]
- Resistance [tex]\(R = 18 \, \Omega\)[/tex]
- Time [tex]\(t = 8.0 \, \text{s}\)[/tex]
Calculate the current:
[tex]\[ I_Z = \frac{12}{18} \approx 0.667 \, \text{A} \][/tex]
Calculate the charge:
[tex]\[ Q_Z = 0.667 \cdot 8.0 \approx 5.336 \, \text{C} \][/tex]
Since [tex]\(5.336 \, \text{C}\)[/tex] is greater than [tex]\(4.0 \, \text{C}\)[/tex], Circuit [tex]\(Z\)[/tex] needs to be reported.
### Conclusion
The circuits that need to be reported are [tex]\(X\)[/tex] and [tex]\(Z\)[/tex]. Therefore, the correct answer is:
[tex]\[ \boxed{X \text{ and } Z} \][/tex]
We appreciate your time on our site. Don't hesitate to return whenever you have more questions or need further clarification. We appreciate your time. Please revisit us for more reliable answers to any questions you may have. Stay curious and keep coming back to Westonci.ca for answers to all your burning questions.
