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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]
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