Question

When a battery is connected to a \(100 .-\Omega\) resistor, the current is \(4.00 \mathrm{~A}\). When the same battery is connected to a \(400 .-\Omega\) resistor, the current is 1.01 A. Find the emf supplied by the battery and the internal resistance of the battery.

Short answer

Based on the given information and calculations, the electromotive force (emf) supplied by the battery is approximately \(1903.19 V\), and the internal resistance of the battery is approximately \(376.19 \Omega\).

Step-by-step solution

Write down the information given in the problem.

We are given the following information: - When connected to a \(100 \Omega\) resistor, the current is \(4.00 A\). - When connected to a \(400 \Omega\) resistor, the current is \(1.01 A\). Our goal is to find the emf supplied by the battery and the internal resistance of the battery, which we will denote by \(E\) and \(r\), respectively.

Use Ohm's law and the equations for internal resistance to set up a system of equations.

First, let's recall Ohm's law, which states that the voltage across a resistor is directly proportional to the current flowing through it: $$V = IR$$ Now, let's consider the equations for the internal resistance of the battery. When a battery is connected to a resistor, the total resistance in the circuit is \(R + r\) (the external resistor plus the internal resistance of the battery). Therefore, the current flowing through the circuit can be written as: $$I = \frac{E}{R + r}$$ Using the given values for the two cases, we can set up a system of equations as follows: Case 1: \(100 \Omega\) resistor and \(4.00 A\) current $$4.00 = \frac{E}{100 + r}$$ Case 2: \(400 \Omega\) resistor and \(1.01 A\) current $$1.01 = \frac{E}{400 + r}$$

Solve the system of equations to find the values of \(E\) and \(r\).

Now, we have two equations with two unknowns, \(E\) and \(r\). We can solve this system of equations using the method of substitution or elimination. In this case, let's use the elimination method. First, we will multiply both equations by suitable factors to make their coefficients equal. Equation 1 multiplied by \(400+r\): $$(4.00)(400 + r) = E$$ Equation 2 multiplied by \(100+r\): $$(1.01)(100 + r) = E$$ Now, since both equations are equal to \(E\), we can set them equal to each other and solve for \(r\): $$(4.00)(400 + r) = (1.01)(100 + r)$$ Expanding and simplifying the equation: $$1600 + 4.00r = 101 + 1.01r$$ Subtract \(1.01r\) and \(101\) from both sides: $$3.99r = 1499$$ Divide by \(3.99\): $$r \approx 376.19 \Omega$$ Now substitute the value of \(r\) back into one of the original equations to find \(E\). Let's use the first equation: $$4.00 = \frac{E}{100 + 376.19}$$ Multiply both sides by \(100 + 376.19\): $$E \approx 1903.19 \mathrm{V}$$

Find the emf and internal resistance of the battery.

From our calculations, the emf supplied by the battery is approximately \(1903.19 V\), and the internal resistance of the battery is approximately \(376.19 \Omega\).

Key concepts

Ohm's Law

Ohm's Law is a fundamental principle in the field of electronics and electrical circuits. It states that the voltage (V) across a resistor in a circuit is directly proportional to the current (I) flowing through it and the resistance (R) of the resistor. This relationship can be expressed with the formula:\[ V = I \times R \]This equation helps us calculate the voltage if we know the current and resistance, or the current if we know the voltage and resistance.
In practical scenarios, Ohm's Law allows us to understand how changing one parameter can affect others, making it extremely useful in electrical engineering and troubleshooting circuits. For example, in the given problem, knowing the resistance and current allowed us to deduce other electrical quantities related to the circuit.

Electromotive Force (EMF)

Electromotive Force, commonly abbreviated as EMF, represents the energy supplied by a battery or power source per unit charge. It is the driving force that pushes electrons around a circuit. In simpler terms, it is the voltage provided by the source of electrical energy when no current is flowing. This doesn't mean it's a force in the traditional sense, rather it is a potential difference, measured in volts (V).
In our exercise:

• The EMF is denoted by \(E\).
• It was calculated using mathematical equations based on varying load resistances and their corresponding currents.

Having an accurate measure of EMF is crucial for assessing how much energy a battery can deliver before internal and external resistances take a toll on its output.

Series Circuit

A series circuit is a type of electrical circuit in which components are connected end-to-end, so there is only one path for current to take. This means every component in the series shares the same current. However, the total resistance of the series circuit is the sum of individual resistances:
\[ R_{total} = R_1 + R_2 + ... + R_n \]In our context:

• The internal resistance of a battery and the resistor it is connected to form a simple series circuit.
• This setup means the total current depends on both external and internal resistances.

Series circuits are quite simple to analyze due to their straightforward nature, but it's important to note that if one part fails, the entire circuit stops working. Understanding these concepts helps in designing and troubleshooting real-life circuits effectively.