Question

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 internal resistance. What is the voltage difference across the second resistor

Answer 1

.737 v

Step-by-step 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