To determine the resistance of a resistor needed in series with a 24V, 120W light bulb when powered by a 60V supply, we can follow these steps:
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Calculate the resistance of the light bulb: The power formula can be used here:
\[ P = \frac{V^2}{R} \] Rearranging this gives:
\[ R = \frac{V^2}{P} \]Substituting the values for the bulb:
- \( V = 24V \)
- \( P = 120W \)
\[ R_{\text{bulb}} = \frac{(24)^2}{120} = \frac{576}{120} = 4.8 , \Omega \]
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Calculate the total resistance needed to drop the extra voltage: The total voltage across the series circuit is 60V. To ensure the 24V light bulb operates correctly, we need to drop the remaining voltage (60V - 24V = 36V) across a resistor in series.
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Current through the light bulb: Calculate the current through the light bulb using: \[ I = \frac{P}{V} \] Substituting the values: \[ I = \frac{120}{24} = 5 , A \]
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Finding the resistance required for the series resistor: We need to find the resistor value such that it drops 36V at a current of 5A. Using Ohm’s law: \[ V = I \times R \] We can rearrange to find \( R \): \[ R = \frac{V}{I} \] Here \( V = 36V \) and \( I = 5A \): \[ R_{\text{series}} = \frac{36}{5} = 7.2 , \Omega \]
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Conclusion: The resistance of the resistor that must be connected in series with the 24V, 120W light bulb to operate correctly from a 60V supply is 7.2 Ω.
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