To determine which circuits deliver more than 4.0 C of charge in 8.0 seconds, we need to calculate the current (I) for each circuit using Ohm's Law and the formula for charge (Q).
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Calculate current using Ohm's Law: \[ I = \frac{V}{R} \] where \( V \) is voltage and \( R \) is resistance.
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Calculate charge using the formula: \[ Q = I \times t \] where \( t \) is time in seconds.
Now, let's calculate the current and charge for each circuit:
Circuit W:
- Voltage \( V = 18 \) volts
- Resistance \( R = 38 \) ohms
- Current \( I = \frac{18}{38} \approx 0.474 \) A
- Charge \( Q = 0.474 , \text{A} \times 8.0 , \text{s} \approx 3.792 , \text{C} \)
Circuit X:
- Voltage \( V = 24 \) volts
- Resistance \( R = 34 \) ohms
- Current \( I = \frac{24}{34} \approx 0.706 \) A
- Charge \( Q = 0.706 , \text{A} \times 8.0 , \text{s} \approx 5.648 , \text{C} \)
Circuit Y:
- Voltage \( V = 34 \) volts
- Resistance \( R = 70 \) ohms
- Current \( I = \frac{34}{70} \approx 0.486 \) A
- Charge \( Q = 0.486 , \text{A} \times 8.0 , \text{s} \approx 3.888 , \text{C} \)
Circuit Z:
- Voltage \( V = 12 \) volts
- Resistance \( R = 18 \) ohms
- Current \( I = \frac{12}{18} \approx 0.667 \) A
- Charge \( Q = 0.667 , \text{A} \times 8.0 , \text{s} \approx 5.336 , \text{C} \)
Now we can summarize the results:
- Circuit W: \( Q \approx 3.792 , \text{C} \) (not included)
- Circuit X: \( Q \approx 5.648 , \text{C} \) (included)
- Circuit Y: \( Q \approx 3.888 , \text{C} \) (not included)
- Circuit Z: \( Q \approx 5.336 , \text{C} \) (included)
Conclusion: The circuits that deliver more than 4.0 C of charge in 8.0 s are circuits X and Z.
Therefore, Bruce must include the circuits X and Z in his report.