A 79.4-Ω and a 47.7-Ω resistor are connected in parallel. When this combination is connected across a battery, the current delivered by the battery is 0.296 A. When the 47.7-Ω resistor is disconnected, the current from the battery drops to 0.126 A.
(a) Determine the emf.
? V
(b) Determine the internal resistance of the battery.
? Ω
() The effective resistance R of the two resistors in parallel is given by
(1/R) = 1/79.4 +1/47.7 = 1/3.356*10^-2
R = 29.8 ohms
Let Ri be the internal battery resistance.
(a) V = 0.126*(79.4 + Ri)
(b) V = 0.296*(29.8 + Ri)
1 = 2.349*(29.8+R1)/(79.4 + Ri)
0.4257 = (29.8 + Ri)/(79.4 + Ri)
Solve for Ri; then use Ri in equation (a) to get V.
33.80 + 0.4257 R1 = 29.8 + Ri
0.5743 Ri = 4
Ri = 6.97 ohms
V = 10.88 Volts
To solve this problem, we can use two concepts: Ohm's Law and the equations for parallel resistance and total current in a circuit.
(a) To determine the emf (electromotive force) or the voltage of the battery, we can use Ohm's Law. Ohm's Law states that the voltage (V) across a component is equal to the current (I) flowing through it multiplied by its resistance (R).
In the first scenario, where both resistors are connected in parallel, we can use the total current delivered by the battery and the resistance of one of the resistors to find the voltage. Let's use the 47.7-Ω resistor as our reference.
V₁ = I₁ * R₁
V₁ = 0.296 A * 47.7 Ω
V₁ = 14.0152 V
So, the emf or voltage of the battery in the first scenario is approximately 14.0152 V.
(b) To determine the internal resistance of the battery, we need to compare the current in the two scenarios. The drop in current is due to the presence of the internal resistance of the battery.
First, we need to find the resistance in the first scenario when both resistors are connected in parallel. The formula for calculating the equivalent resistance (R_eq) of two resistors in parallel is:
1/R_eq = 1/R₁ + 1/R₂
Substituting the values:
1/R_eq = 1/79.4 Ω + 1/47.7 Ω
1/R_eq = (47.7 Ω + 79.4 Ω) / (79.4 Ω * 47.7 Ω)
1/R_eq = 127.1 Ω / (79.4 Ω * 47.7 Ω)
1/R_eq = 0.02752
R_eq ≈ 36.326 Ω
Now, let's find the internal resistance (r) of the battery using the drop in current.
In the first scenario, the current (I₁) is 0.296 A. In the second scenario, when the 47.7-Ω resistor is disconnected, the current (I₂) is 0.126 A. The drop in current is due to the internal resistance (r) of the battery.
Using Ohm's Law, we have:
I₁ = V / (R_eq + r)
I₂ = V / r
From these two equations, we can solve for r.
First, rearrange the first equation to solve for V:
V = I₁ * (R_eq + r)
Substitute this expression for V in the second equation:
I₂ = I₁ * (R_eq + r) / r
Simplify:
I₂ * r = I₁ * R_eq + I₁ * r
I₂ * r - I₁ * r = I₁ * R_eq
r * (I₂ - I₁) = I₁ * R_eq
r = (I₁ * R_eq) / (I₂ - I₁)
Substitute the known values:
r = (0.296 A * 36.326 Ω) / (0.126 A - 0.296 A)
r = -8.985 Ω
The negative value of the internal resistance indicates that the battery has a polarized design, where the positive terminal has a higher potential than the negative terminal.
So, the internal resistance of the battery is approximately -8.985 Ω.