Answer:
The resistance of each resistor is 2.5 Ω
The potential difference across each resistor is 2.4 V.
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)
[tex]R = \dfrac{12}{0.961}\,\Omega[/tex]
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
[tex]= \dfrac{12}{5\times0.961}\,\Omega = 2.5\,\Omega[/tex]
Also, because they are in series and are equal, the EMF is distributed across them equally. Therefore, the potential difference across each resistor is
[tex]\dfrac{12.0\text{ V}}{5} = 2.40\text{ V}[/tex]
Answer:
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
