Question

A battery with an emf of 12.0 V shows a terminal voltage of 11.7 V when operating in a circuit with two lightbulbs, each rated at 4.0 W (at 12.0 V), which are connected in parallel.
What is the battery's internal resistance?

Answer 1

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 = $\frac{V^{2} }{R}$

=> R = $\frac{V^{2} }{P}$             -------------------(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 = $\frac{V^{2} }{P}$    --------------------------------(iii)

Where;

V = 12.0V

P = 4.0W

Substitute these values into equation (iii) as follows;

R₁ = $\frac{12.0^{2} }{4}$

R₁ = $\frac{144}{4}$

R₁ = 36Ω

Following the same approach, to get R₂, equation (ii) can be written as;

R₂ = $\frac{V^{2} }{P}$    --------------------------------(iv)

Where;

V = 12.0V

P = 4.0W

Substitute these values into equation (iv) as follows;

R₂ = $\frac{12.0^{2} }{4}$

R₂ = $\frac{144}{4}$

R₂ = 36Ω

Now, since the bulbs are connected in parallel, the effective resistance (Rₓ) is given by;

$\frac{1}{R_{X} }$ = $\frac{1}{R_1}$ + $\frac{1}{R_2}$       -----------------(v)

Substitute the values of R₁ and R₂ into equation (v) as follows;

$\frac{1}{R_X}$ = $\frac{1}{36}$ + $\frac{1}{36}$

$\frac{1}{R_X}$ = $\frac{2}{36}$

Rₓ = $\frac{36}{2}$

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 = $\frac{11.7}{18}$

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 = $\frac{0.3}{0.65}$

r = 0.46Ω

Therefore, the internal resistance of the battery is 0.46Ω

Answer 2

Answer:

$R_i_n_t=0.45 \Omega$

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:

$R_i_n_t=(\frac{V_N_L}{V_F_L} -1)R_L$

Where:

$V_F_L=Load\hspace{3}voltage=11.7V\\V_N_L= O pen\hspace{3}circuit\hspace{3}voltage=12V\\R_L=Load\hspace{3}resistance$

As you can see, we don't know the exactly value of the $R_L$. 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:

$\frac{12}{11.7} =\frac{4}{P}$

Solving for $P$ :

$P=\frac{11.7*4}{12} =3.9W$

Now, the electric power is given by:

$P=\frac{V^2}{R_b}$

Where:

$R_b=Resistance\hspace{3}of\hspace{3}each\hspace{3}lightbulb$

So:

$R_b=\frac{V^2}{P} =\frac{11.7^2}{3.9} =35.1\Omega$

Now, because of the lightbulbs are connected in parallel the equivalent resistance is given by:

$\frac{1}{R_L} =\frac{1}{R_b} +\frac{1}{R_b} =\frac{2}{R_b} \\\\ R_L= \frac{R_b}{2} =\frac{35.1}{2}=17.55\Omega$

Finally, now we have all the data, let's replace it into the internal resistance equation:

$R_i_n_t=(\frac{12}{11.7} -1)17.55=0.45\Omega$