You connect five identical resistors in series to a battery whose EMF is 12.0 V and whose internal resistance is negligible. You measure the current that the circuit draws from the battery and find 0.961 A. What are the resistance of each resistor and the potential difference across each resistor?

R = 2.4974 Ω

V = 2.4 V

Explanation:

When Five identical resistors are connected in series,

R' = R+R+R+R+R

R' = 5R Equation 1

Where R' = Combined resistance of the five resistors connected in series, R = Resistance of each of the resistor.

make R the subject of the equation

R = R'/5 Equation 2

Using

E = I(R'+r) Equation 3

Where E = Emf of the battery, I = current, r = internal resistance of the battery

Given: E = 12 V, I = 0.961 A, r = 0 Ω (negligible)

Substitute into equation 3

12 = 0.961(R')

R' = 12/0.961

R' = 12.487 Ω

Substitute into equation 2

R = 12.487/5

R = 2.4974 Ω

using ohm's law,

V = IR Equation 4

Where V = potential difference across each resistance.

Given: I = 0.961 A, R = 2.4974 Ω

Substitute into equation 4

V = 0.961(2.4974)

V = 2.4 V.

Hence the resistance and potential difference across each resistor = 2.4974 Ω  and 2.4 V

305.7

explanation:

density =m/v

density =917/3.00

density = 305.

approximately = 305.7

in four significant figures