Jackie rides her bike to school every day and enjoys playing beach volleyball on weekends. She also helps with tasks around the
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<h2>Answer:</h2>
0.46Ω
<h2>Explanation:</h2>The electromotive force (E) in the circuit is related to the terminal voltage(V), of the circuit and the internal resistance (r) of the battery as follows;
E = V + Ir --------------------(a)
Where;
I = current flowing through the circuit
But;
V = I x Rₓ ---------------------(b)
Where;
Rₓ = effective or total resistance in the circuit.
<em>First, let's calculate the effective resistance in the circuit:</em>
The effective resistance (Rₓ) in the circuit is the one due to the resistances in the two lightbulbs.
Let;
R₁ = resistance in the first bulb
R₂ = resistance in the second bulb
Since the two bulbs are both rated at 4.0W ( at 12.0V), their resistance values (R₁ and R₂) are the same and will be given by the power formula;
P =
=> R = -------------------(ii)
Where;
P = Power of the bulb
V = voltage across the bulb
R = resistance of the bulb
To get R₁, equation (ii) can be written as;
R₁ = --------------------------------(iii)
Where;
V = 12.0V
P = 4.0W
Substitute these values into equation (iii) as follows;
R₁ =
R₁ =
R₁ = 36Ω
Following the same approach, to get R₂, equation (ii) can be written as;
R₂ = --------------------------------(iv)
Where;
V = 12.0V
P = 4.0W
Substitute these values into equation (iv) as follows;
R₂ =
R₂ =
R₂ = 36Ω
Now, since the bulbs are connected in parallel, the effective resistance (Rₓ) is given by;
=
+
-----------------(v)
Substitute the values of R₁ and R₂ into equation (v) as follows;
=
+
=
Rₓ =
Rₓ = 18Ω
The effective resistance (Rₓ) is therefore, 18Ω
<em>Now calculate the current I, flowing in the circuit:</em>
Substitute the values of V = 11.7V and Rₓ = 18Ω into equation (b) as follows;
11.7 = I x 18
I =
I = 0.65A
<em>Now calculate the battery's internal resistance:</em>
Substitute the values of E = 12.0, V = 11.7V and I = 0.65A into equation (a) as follows;
12.0 = 11.7 + 0.65r
0.65r = 12.0 - 11.7
0.65r = 0.3
r =
r = 0.46Ω
Therefore, the internal resistance of the battery is 0.46Ω
The concept required to solve this problem is linked to inductance. This can be defined as the product between the permeability in free space by the number of turns squared by the area over the length. Recall that Inductance is defined as the opposition of a conductive element to changes in the current flowing through it. Mathematically it can be described as
Here,
= Permeability at free space
N = Number of loops
A = Cross-sectional Area
l = Length
Replacing with our values we have,
Therefore the Inductance is