How much resistance does a heavy duty flashlight have if it has a current of 25 mA flowing through it and is being
powered by four 1.5 V cells?
Answers
Answered by
7
Answer:
Explanation:
CURRENT(I)=25x 10^-3 A
POTENTIAL DIFFERENCE=1.5V
RESISTANCE(R)=V/I
=1.5/25x10^-3
1.5x10^3/25
=0.06x10^3
=6 ohm
Answered by
8
Answer:
240ohms
Explanation:
V=IR
V=6v. (there are 4 1.5 v cells in series, as is the case with flashlights anyway)
25mA=0.025A.
6=0.025(R)--> 6/0.025=R--> 240Ohms
continue to POWER... P=IV --> P=0.025(6)--> P=0.15w
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