Respuesta :
Answer:
0.46Ω
Explanation:
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.
First, let's calculate the effective resistance in the circuit:
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 = [tex]\frac{V^{2} }{R}[/tex]
=> R = [tex]\frac{V^{2} }{P}[/tex] -------------------(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 = [tex]\frac{V^{2} }{P}[/tex] --------------------------------(iii)
Where;
V = 12.0V
P = 4.0W
Substitute these values into equation (iii) as follows;
R₁ = [tex]\frac{12.0^{2} }{4}[/tex]
R₁ = [tex]\frac{144}{4}[/tex]
R₁ = 36Ω
Following the same approach, to get R₂, equation (ii) can be written as;
R₂ = [tex]\frac{V^{2} }{P}[/tex] --------------------------------(iv)
Where;
V = 12.0V
P = 4.0W
Substitute these values into equation (iv) as follows;
R₂ = [tex]\frac{12.0^{2} }{4}[/tex]
R₂ = [tex]\frac{144}{4}[/tex]
R₂ = 36Ω
Now, since the bulbs are connected in parallel, the effective resistance (Rₓ) is given by;
[tex]\frac{1}{R_{X} }[/tex] = [tex]\frac{1}{R_1}[/tex] + [tex]\frac{1}{R_2}[/tex] -----------------(v)
Substitute the values of R₁ and R₂ into equation (v) as follows;
[tex]\frac{1}{R_X}[/tex] = [tex]\frac{1}{36}[/tex] + [tex]\frac{1}{36}[/tex]
[tex]\frac{1}{R_X}[/tex] = [tex]\frac{2}{36}[/tex]
Rₓ = [tex]\frac{36}{2}[/tex]
Rₓ = 18Ω
The effective resistance (Rₓ) is therefore, 18Ω
Now calculate the current I, flowing in the circuit:
Substitute the values of V = 11.7V and Rₓ = 18Ω into equation (b) as follows;
11.7 = I x 18
I = [tex]\frac{11.7}{18}[/tex]
I = 0.65A
Now calculate the battery's internal resistance:
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 = [tex]\frac{0.3}{0.65}[/tex]
r = 0.46Ω
Therefore, the internal resistance of the battery is 0.46Ω
Answer:
[tex]R_i_n_t=0.45 \Omega[/tex]
Explanation:
Internal resistance is a concept that helps model the electrical consequences of the complex chemical reactions that occur within a battery. When a charge is applied to a battery, the internal resistance can be calculated using the following equation:
[tex]R_i_n_t=(\frac{V_N_L}{V_F_L} -1)R_L[/tex]
Where:
[tex]V_F_L=Load\hspace{3}voltage=11.7V\\V_N_L= O pen\hspace{3}circuit\hspace{3}voltage=12V\\R_L=Load\hspace{3}resistance[/tex]
As you can see, we don't know the exactly value of the [tex]R_L[/tex]. However we can calculated that value using the next simple operations:
The problem tell us that the power of each lightbulb is 4.0 W at 12.0 V, hence let's calculated the power at 11.7V using Cross-multiplication:
[tex]\frac{12}{11.7} =\frac{4}{P}[/tex]
Solving for [tex]P[/tex] :
[tex]P=\frac{11.7*4}{12} =3.9W[/tex]
Now, the electric power is given by:
[tex]P=\frac{V^2}{R_b}[/tex]
Where:
[tex]R_b=Resistance\hspace{3}of\hspace{3}each\hspace{3}lightbulb[/tex]
So:
[tex]R_b=\frac{V^2}{P} =\frac{11.7^2}{3.9} =35.1\Omega[/tex]
Now, because of the lightbulbs are connected in parallel the equivalent resistance is given by:
[tex]\frac{1}{R_L} =\frac{1}{R_b} +\frac{1}{R_b} =\frac{2}{R_b} \\\\ R_L= \frac{R_b}{2} =\frac{35.1}{2}=17.55\Omega[/tex]
Finally, now we have all the data, let's replace it into the internal resistance equation:
[tex]R_i_n_t=(\frac{12}{11.7} -1)17.55=0.45\Omega[/tex]
