Four resistors (67 ohm, 83 ohm, 433 ohm, and 309 ohm in that order) are connected in series to a 7.92 V battery of negligible in
ternal resistance. What is the voltage difference across the second resistor
1 answer:
Answer:
.737 v
Explanation:
Since they are in series....they all have the same current running through them.....find the total resistance to calculate the current:
R = 67 + 83 + 433 + 309 = 892 ohm
V/R = current = 7.92 / 892 = 8.87 mAmps
Now the voltage across ecah resistor is I R
for the second one 8.87 ma * 83 ohm = V = .737 V
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