Question

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?

Answer 1

R = 2.4974 Ω

V = 2.4 V

Step-by-step 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

Answer 2

The resistance of each resistor is 2.5 Ω

The potential difference across each resistor is 2.4 V.

Step-by-step explanation:

By Ohm's law,

V = IR

where V is the potential difference or voltage across an element, I is the current flowing through it and R is its effective resistance.

For the group of five resistors, let their combined resistance be R.

Then

(12.0 V) = (0.961 A)(R)

`R = (12)/(0.961)\,\Omega`

Because they are in series, R is the arithmetic sum of their individual resistances. Because they are all identical, the resistance of each resistor is

`= (12)/(5*0.961)\,\Omega = 2.5\,\Omega`

Also, because they are in series and are equal, the EMF is distributed across them equally. Therefore, the potential difference across each resistor is

`\frac{12.0\text{ V}}{5} = 2.40\text{ V}`