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2 years ago

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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:

Whitepunk [10]2 years ago

7 0

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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From the question we are told that

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