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Sagot :
To solve this problem, we'll need to follow a sequence of steps involving the principles of electric circuits and the mathematical relationship between voltage, resistance, current, and charge.
1. Understanding the Problem:
We are given:
- Voltage (V) and Resistance (R) for each circuit.
- A time (t) of 8.0 seconds.
- A charge threshold of 4.0 Coulombs (C).
2. Key Equations:
- Ohm's Law: [tex]\( I = \frac{V}{R} \)[/tex]
This equation relates the current (I) flowing through a circuit to its voltage (V) and resistance (R).
- Charge Calculation: [tex]\( Q = I \times t \)[/tex]
This equation finds the charge (Q) delivered by a current (I) over time (t).
3. Procedure:
- We will use Ohm's Law to find the current for each circuit.
- Next, we'll calculate the charge delivered in 8.0 seconds.
- Finally, we'll compare the calculated charge for each circuit to the threshold of 4.0 C to determine which circuits deliver more than 4.0 C of charge.
Let's apply these steps to each circuit.
Circuit W:
- Voltage (V): 18 V
- Resistance (R): 38 Ω
Calculate current:
[tex]\[ I_W = \frac{V_W}{R_W} = \frac{18}{38} \approx 0.474 \text{ A} \][/tex]
Calculate charge:
[tex]\[ Q_W = I_W \times t = 0.474 \, \text{A} \times 8 \, \text{s} \approx 3.792 \, \text{C} \][/tex]
Circuit X:
- Voltage (V): 24 V
- Resistance (R): 34 Ω
Calculate current:
[tex]\[ I_X = \frac{V_X}{R_X} = \frac{24}{34} \approx 0.706 \text{ A} \][/tex]
Calculate charge:
[tex]\[ Q_X = I_X \times t = 0.706 \, \text{A} \times 8 \, \text{s} \approx 5.648 \, \text{C} \][/tex]
Circuit Y:
- Voltage (V): 34 V
- Resistance (R): 70 Ω
Calculate current:
[tex]\[ I_Y = \frac{V_Y}{R_Y} = \frac{34}{70} \approx 0.486 \text{ A} \][/tex]
Calculate charge:
[tex]\[ Q_Y = I_Y \times t = 0.486 \, \text{A} \times 8 \, \text{s} \approx 3.888 \, \text{C} \][/tex]
Circuit Z:
- Voltage (V): 12 V
- Resistance (R): 18 Ω
Calculate current:
[tex]\[ I_Z = \frac{V_Z}{R_Z} = \frac{12}{18} \approx 0.667 \text{ A} \][/tex]
Calculate charge:
[tex]\[ Q_Z = I_Z \times t = 0.667 \, \text{A} \times 8 \, \text{s} \approx 5.336 \, \text{C} \][/tex]
Summary of Charges:
- [tex]\( Q_W \approx 3.792 \, \text{C} \)[/tex]
- [tex]\( Q_X \approx 5.648 \, \text{C} \)[/tex]
- [tex]\( Q_Y \approx 3.888 \, \text{C} \)[/tex]
- [tex]\( Q_Z \approx 5.336 \, \text{C} \)[/tex]
Comparing these values to the threshold of 4.0 C, we can see that:
- Circuits [tex]\(X\)[/tex] and [tex]\(Z\)[/tex] deliver more than 4.0 C of charge.
Therefore, Bruce must include circuits [tex]\(X\)[/tex] and [tex]\(Z\)[/tex] in his report.
So the answer is: [tex]\(X\)[/tex] and [tex]\(Z\)[/tex].
1. Understanding the Problem:
We are given:
- Voltage (V) and Resistance (R) for each circuit.
- A time (t) of 8.0 seconds.
- A charge threshold of 4.0 Coulombs (C).
2. Key Equations:
- Ohm's Law: [tex]\( I = \frac{V}{R} \)[/tex]
This equation relates the current (I) flowing through a circuit to its voltage (V) and resistance (R).
- Charge Calculation: [tex]\( Q = I \times t \)[/tex]
This equation finds the charge (Q) delivered by a current (I) over time (t).
3. Procedure:
- We will use Ohm's Law to find the current for each circuit.
- Next, we'll calculate the charge delivered in 8.0 seconds.
- Finally, we'll compare the calculated charge for each circuit to the threshold of 4.0 C to determine which circuits deliver more than 4.0 C of charge.
Let's apply these steps to each circuit.
Circuit W:
- Voltage (V): 18 V
- Resistance (R): 38 Ω
Calculate current:
[tex]\[ I_W = \frac{V_W}{R_W} = \frac{18}{38} \approx 0.474 \text{ A} \][/tex]
Calculate charge:
[tex]\[ Q_W = I_W \times t = 0.474 \, \text{A} \times 8 \, \text{s} \approx 3.792 \, \text{C} \][/tex]
Circuit X:
- Voltage (V): 24 V
- Resistance (R): 34 Ω
Calculate current:
[tex]\[ I_X = \frac{V_X}{R_X} = \frac{24}{34} \approx 0.706 \text{ A} \][/tex]
Calculate charge:
[tex]\[ Q_X = I_X \times t = 0.706 \, \text{A} \times 8 \, \text{s} \approx 5.648 \, \text{C} \][/tex]
Circuit Y:
- Voltage (V): 34 V
- Resistance (R): 70 Ω
Calculate current:
[tex]\[ I_Y = \frac{V_Y}{R_Y} = \frac{34}{70} \approx 0.486 \text{ A} \][/tex]
Calculate charge:
[tex]\[ Q_Y = I_Y \times t = 0.486 \, \text{A} \times 8 \, \text{s} \approx 3.888 \, \text{C} \][/tex]
Circuit Z:
- Voltage (V): 12 V
- Resistance (R): 18 Ω
Calculate current:
[tex]\[ I_Z = \frac{V_Z}{R_Z} = \frac{12}{18} \approx 0.667 \text{ A} \][/tex]
Calculate charge:
[tex]\[ Q_Z = I_Z \times t = 0.667 \, \text{A} \times 8 \, \text{s} \approx 5.336 \, \text{C} \][/tex]
Summary of Charges:
- [tex]\( Q_W \approx 3.792 \, \text{C} \)[/tex]
- [tex]\( Q_X \approx 5.648 \, \text{C} \)[/tex]
- [tex]\( Q_Y \approx 3.888 \, \text{C} \)[/tex]
- [tex]\( Q_Z \approx 5.336 \, \text{C} \)[/tex]
Comparing these values to the threshold of 4.0 C, we can see that:
- Circuits [tex]\(X\)[/tex] and [tex]\(Z\)[/tex] deliver more than 4.0 C of charge.
Therefore, Bruce must include circuits [tex]\(X\)[/tex] and [tex]\(Z\)[/tex] in his report.
So the answer is: [tex]\(X\)[/tex] and [tex]\(Z\)[/tex].
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