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If a 6Ω resistor is connected to a 9V battery, what is the power dissipated by t
If a 6Ω resistor is connected to a 9V battery, what is the power dissipated by the resistor?
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Q1
If a 6Ω resistor is connected to a 9V battery, what is the power dissipated by the resistor?
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Power P = V^2 / R = (9V)^2 / 6Ω = 81 / 6 = 13.5W.
Questions & Step-by-step Solutions
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Q
Q: If a 6Ω resistor is connected to a 9V battery, what is the power dissipated by the resistor?
Solution:
Power P = V^2 / R = (9V)^2 / 6Ω = 81 / 6 = 13.5W.
Steps: 7
Show Steps
Step 1: Identify the values given in the problem. We have a resistor with a resistance of 6 ohms (Ω) and a battery with a voltage of 9 volts (V).
Step 2: Recall the formula for power (P) in terms of voltage (V) and resistance (R). The formula is P = V^2 / R.
Step 3: Substitute the values into the formula. Replace V with 9V and R with 6Ω: P = (9V)^2 / 6Ω.
Step 4: Calculate (9V)^2. This equals 81V^2.
Step 5: Now divide 81V^2 by 6Ω: 81 / 6.
Step 6: Perform the division. 81 divided by 6 equals 13.5.
Step 7: Therefore, the power dissipated by the resistor is 13.5 watts (W).
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