91Ó°ÊÓ

Ohm’s Law is a fundamental principle in electric circuits. It establishes a simple relationship between voltage, current, and resistance. The law is succinctly expressed as \( V = IR \), where:

• \( V \) is the voltage (in volts).
• \( I \) is the current (in amperes).
• \( R \) is the resistance (in ohms).

This equation implies that if you know any two of these values, you can easily calculate the third.
For example, in this problem, we use Ohm’s Law to find the total current flowing from the battery. Knowing the total resistance of the circuit and the voltage, we can rearrange the equation to \( I = \frac{V}{R} \) to find the current.
Understanding Ohm’s Law is crucial because it helps you predict how circuits behave under different conditions.
This becomes particularly helpful for troubleshooting and designing circuits efficiently.

Understanding how resistors work in series and parallel is key to analyzing electric circuits.
When resistors are connected in series, the total resistance is just the sum of individual resistances. This is because the current must pass through each resistor one after the other.In our scenario, the 20-ohm resistor is in series with the parallel group.

• Formula: \( R_{\text{total}} = R_1 + R_2 + \ldots + R_n \).
• Total resistance for series: \( R_{\text{series}} = 40.0\,\Omega + 20.0\,\Omega = 60.0\,\Omega \).

On the other hand, resistors in parallel provide multiple pathways for the current.
This leads to a lower overall resistance than any single resistor in the network. You calculate parallel resistances using the equation:

• \( \frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} + \ldots + \frac{1}{R_n} \).
• Here, \( R_{\text{parallel}} = 40.0\,\Omega \), calculated using the ones given: \( 60.0\,\Omega \) and \( 120.0\,\Omega \).

Mastering these two configurations enables you to evaluate complex circuits much more easily.

Power calculations determine the energy consumed or delivered by components within a circuit. Power (\( P \)) is the rate at which energy is used and is measured in watts (W).
Power can be calculated using various equations, each involving a different set of known quantities. For our problem, we focus on:

• \( P = IV \), where \( I \) is the current and \( V \) is the voltage.
• Alternatively, \( P = I^2R \), involving current and resistance.
• \( P = \frac{V^2}{R} \), when only voltage and resistance are known.

In this exercise, the power delivered to the \( 120.0\,\Omega \) resistor is found using \( P = IV \).
With a calculated current of \( 0.0833\,\text{A} \) and a voltage of \( 10.0\,\text{V} \), the power is \( P = 0.833\,\text{W} \).
These calculations are essential for understanding how much energy is being converted in a circuit, identifying efficiency, and avoiding overheating in components.