A flashlight uses batteries that add up to 6.0 V and has a power output of 0.41 W.
How much current is drawn from the batteries?
____A
What is the effective resistance of the flashlight?
____Ω
To find the current drawn from the batteries, we can use the formula:
Power (P) = Voltage (V) * Current (I)
0.41 W = 6.0 V * I
I = 0.41 W / 6.0 V
I = 0.0683 A
Therefore, the current drawn from the batteries is 0.0683 A.
To find the effective resistance of the flashlight, we can use Ohm's Law:
Resistance (R) = Voltage (V) / Current (I)
R = 6.0 V / 0.0683 A
R = 87.8 Ω
Therefore, the effective resistance of the flashlight is 87.8 Ω.