Problem 40 - Physics

Question

A 60.0 -W lamp is placed in series with a resistor and a 120.0 -V source. If the voltage across the lamp is \(25 \mathrm{~V},\) what is the resistance \(R\) of the resistor?

Step-by-Step Solution

Answer

The resistance of the resistor is approximately 39.58 ohms.

Step 1: Understand the Problem

We have a circuit with a lamp and a resistor in series connected to a 120 V source. The voltage across the lamp is 25 V, and we need to find the resistance of the resistor.

Step 2: Calculate Current through the Lamp

Since the lamp is specified to be 60 W, we can find the current through the lamp using the power formula: \[ P = V imes I. \] Plugging in the values, we have: \[ 60 = 25 imes I. \] Solving for \( I \), we find: \[ I = \frac{60}{25} = 2.4 \text{ A}. \]

Step 3: Determine the Voltage across the Resistor

Since the total voltage of the circuit is 120 V and the voltage across the lamp is 25 V, the voltage across the resistor can be found using:\[ V_{\text{resistor}} = V_{\text{total}} - V_{\text{lamp}} = 120 - 25 = 95 \text{ V}. \]

Step 4: Calculate the Resistance of the Resistor

Now that we know the voltage across the resistor is 95 V and the current is 2.4 A (the same as through the lamp because they are in series), use Ohm’s Law to find the resistance:\[ R = \frac{V}{I} = \frac{95}{2.4} \approx 39.58 \text{ ohms}. \]

Key Concepts

Ohm's Law

At the heart of circuit analysis is Ohm's Law, a fundamental concept crucial for understanding electrical circuits. It defines the relationship between voltage, current, and resistance in an electrical circuit. The law is expressed by the formula \( V = I \times R \), where:

• \(V\) stands for voltage, measured in volts (V),
• \(I\) represents current, measured in amperes (A),
• \(R\) is resistance, measured in ohms (Ω).

Ohm's Law shows that the voltage across a component in a circuit is the product of the current flowing through it and its resistance.
This simple formula helps us decipher many aspects of a circuit's behavior, including determining unknown values once two variables are known. For example, if you know both the current and resistance, you can easily calculate the voltage across a resistor. This concept is widely applicable in engineering and electronics, forming the basis for designing and analyzing electrical systems.
In our exercise, Ohm's Law was key in finding the resistance of an unknown resistor once we knew the current and the voltage across it.

Series Circuit

A series circuit is one of the basic types of electrical circuits where components are linked one after the other. In this configuration, the same electric current flows through each component.
This unique arrangement leads to several important properties:

• The total voltage across a series circuit is equal to the sum of the individual voltages across each component.
• Since the current is the same throughout, Ohm's Law can be applied repeatedly across each series component to find unknown values.
• If one element fails, the entire circuit is disrupted because there is only one path for the current.

In our problem, a lamp and a resistor are in series. The total voltage is shared between the lamp and the resistor. Knowing the voltage across the lamp and the total voltage allowed us to determine the voltage across the resistor. This understanding of series circuits leads us to accurately perform calculations needed to analyze the circuit.

Resistor Calculation

Resistors are common components used to control current and voltage within a circuit. Calculating the resistance is important for circuit design and analysis. To find the resistance, we often rely on Ohm's Law.
In step 4 of the solution, we used the known values of voltage across the resistor and the current flowing through it to find the resistance using the formula \( R = \frac{V}{I} \).
This calculation is straightforward; however, accurately identifying and using the given values is crucial.
For a clear resistor calculation:

• Ensure you have the correct voltage across the resistor; in our series circuit, it was the total voltage minus the lamp’s voltage.
• Use the current that flows through the component, which in a series is the same through all elements.
• Double check units to ensure they are compatible for calculations.

By applying these techniques, we arrived at a precise resistance of approximately 39.58 ohms in our problem. This illustrates how resistors are analyzed in practical circuit applications.