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Chapter 24: Problem 29

A 4.5 -W flashlight bulb draws 750 mA. (a) At what voltage does it operate? (b) What's its resistance?

Short Answer

Expert verified
a) The operating voltage of the bulb is 6 V. b) The resistance of the bulb is 8 \, \Omega .

Step by step solution

01

Calculate Voltage

The power \( P \) provided is 4.5 W and the current \( I \) is 750 mA (milliampere), but it needs to be converted to ampere (A) as 1 A = 1000 mA. So, 750 mA becomes 0.75 A. We plug these values into the power formula or the rearranged version of the power formula \( V = \frac{P}{I} \) to find the voltage \( V \). Thus, \( V = \frac{4.5 W}{0.75 A} = 6 V \).
02

Calculate Resistance

Using the calculated voltage \( V = 6 V \) and the current \( I = 0.75 A \) from step 1, we can now use Ohm's law to calculate resistance \( R \). So, \( R = \frac{V}{I} = \frac{6 V}{0.75 A} = 8 \, \Omega \).

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