A flashlight bulb is connected to a dry cell of voltage 2.25 V. It draws 35.0 mA (1000 mA = 1 A). What is its resistance?
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To calculate the resistance of the flashlight bulb, we can use Ohm's Law, which states:
\[ V = I \times R \]
where: - \( V \) is the voltage across the bulb (in volts, V) - \( I \) is the current through the bulb (in amperes, A) - \( R \) is the resistance of the bulb (in ohms, Ω)
Given in the problem: - The voltage (\( V \)) is 2.25 V - The current (\( I \)) is 35.0 mA, which is equivalent to 35.0/1000 A (since 1000 mA = 1 A), that is 0.035 A
We need to solve for \( R \) (resistance):
\[ R = \frac{V}{I} \]
Plug in the values:
\[ R = \frac{2.25 \text{ V}}{0.035 \text{ A}} \] \[ R = 64.2857... \text{ Ω} \]
So the resistance of the flashlight bulb is approximately 64.3 Ω (ohms, rounded to one decimal place).