Question

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

Short answer

Answer: The EMF supplied by the battery is approximately 405.36 V, and the internal resistance of the battery is approximately 1.34 Ω.

Step-by-step solution

1. Assign Variables

Let the emf supplied by the battery be E volts and the internal resistance of the battery be r Ω.

2. Write Ohm's Law and Kirchhoff's Loop Rule for 100 Ω Resistor

We know that the voltage supplied by the battery is equal to the voltage drops across the resistor and the internal resistance of the battery. We can write this as E = I_1 * R_1 + I_1 * r where I_1 is the current through the 100 Ω resistor (4.00 A) and R_1 is the resistance of the 100 Ω resistor.

3. Write Ohm's Law and Kirchhoff's Loop Rule for 400 Ω Resistor

Similarly, for the 400 Ω resistor, we can write the equation as E = I_2 * R_2 + I_2 * r where I_2 is the current through the 400 Ω resistor (1.01 A) and R_2 is the resistance of the 400 Ω resistor.

4. Solve Equations

Now we have two equations: E = 4.00 * 100 + 4.00 * r (1) E = 1.01 * 400 + 1.01 * r (2) From equation (1), we get: E = 400 + 4r (1') From equation (2), we get: E = 404 + 1.01r (2') Subtract equation (1') from (2'): 0 = 4 - 2.99r Now, we solve for r: 2.99r = 4 r = \frac{4}{2.99} r ≈ 1.34 Ω

5. Find EMF

Now that we have the internal resistance, we can substitute r in equation (1') to find the emf (E): E = 400 + 4 * 1.34 E ≈ 405.36 V The emf supplied by the battery is E ≈ 405.36 V and the internal resistance of the battery is r ≈ 1.34 Ω.

Key concepts

Ohm's Law

Ohm's Law is a fundamental principle in the field of electronics and electrical engineering that relates the voltage (V), current (I), and resistance (R) in an electrical circuit. It states that the current flowing through a conductor between two points is directly proportional to the voltage across the two points, and inversely proportional to the resistance between them. Mathematically, this relationship is expressed as: \( V = I \times R \).

In simple terms, if you increase the voltage while keeping the resistance constant, the current will increase accordingly. Conversely, if you increase the resistance while keeping the voltage constant, the current will decrease. Ohm's Law is used to calculate one of these quantities if the other two are known and forms the basis for deriving more complex electrical laws, like Kirchhoff's Loop Rule.

This law becomes particularly helpful in solving our textbook problem, as it allows us to relate the electromotive force (emf) of the battery to the current through, and the resistance of, the separate resistive elements within the circuit.

Kirchhoff's Loop Rule

Kirchhoff's Loop Rule is a fundamental tool in circuit analysis, associated with Kirchhoff's circuit laws, which also include the Current Law (or Junction Rule). Kirchhoff's Loop Rule states that the sum of the potential differences (voltage) in a closed circuit loop must equal zero. This is because a loop is a closed path, and no energy is lost as charge moves around it; energy supplied by any emf sources like batteries must be equal to the energy dissipated through resistors and other components.

Mathematically, Kirchhoff's Loop Rule is written as: \( \text{emf} - (I \times R_{total}) - (I \times r) = 0 \), where \( R_{total} \) is the total resistance, and \( r \) is the internal resistance of the emf source. By applying the Loop Rule, we set up equations that correlate the battery's emf with the current through different resistors, which allows us to calculate the internal resistance and ultimately, calculate the emf itself, as we did in the step by step solution provided.

Internal Resistance

Internal resistance, often denoted by the lowercase letter \( r \), is an inherent property of batteries and other emf sources that causes them to resist the flow of electric current internally. When a current flows through a battery, a small amount of voltage is lost inside the battery due to its internal resistance. This means that the terminal voltage of the battery (the voltage across its terminals) is slightly less than the emf of the battery.

The effect of internal resistance becomes important in our exercise as it affects the voltage output from the battery and the current measured in the circuit. When calculating the battery's emf using Ohm's Law and Kirchhoff's Loop Rule, we must account for the voltage drop caused by the internal resistance, as seen in the steps of the provided solution. This was precisely the approach used to calculate the battery’s internal resistance before determining the emf.

Resistor

A resistor is a passive electrical component that implements electrical resistance as a circuit element. Resistors are used to control the flow of electric current by providing a certain amount of resistance. They come in various values, allowing for precise control over circuit functions, such as adjusting signal levels, dividing voltages, biasing active elements, and limiting current.

In our textbook solution, two different resistors, namely a \( 100 \text{-} \(\Omega\) \) and a \( 400 \text{-} \(\Omega\) \) resistor, are used to vary the current flowing from the battery. This creates two sets of conditions, enabling the use of Ohm's Law and Kirchhoff’s Loop Rule to set up two distinct equations. Solving these equations simultaneously provides the values for both the emf and internal resistance of the battery. In practice, resistors have a wide range of applications, from small electronic projects to large-scale industrial designs.