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Ohm's Law is a basic principle in electronics that helps us understand the relationship between voltage, current, and resistance. It's expressed with the formula \( V = IR \).
In this formula:

• \( V \) represents voltage (in volts). It is the potential difference that makes electric charges move.
• \( I \) denotes current (in amperes). It measures the flow of electric charge.
• \( R \) stands for resistance (in ohms). It indicates how much a material resists the flow of current.

In a parallel circuit, like in this exercise, the voltage across each component is equal. Therefore, when calculating the current through each resistor using Ohm’s Law, we use the formula \( I = \frac{V}{R} \).
This shows how current varies with different resistances while keeping voltage constant. In simpler terms, if you know two of these values, you can find the third!

In circuits, equivalent resistance is the single resistance that can replace a combination of resistors, giving the same effect on current flow. For resistors in parallel, calculating equivalent resistance is vital to simplifying the circuit analysis. The formula used here is:\[\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}\]Where \( R_1, R_2, R_3 \) are the resistances of different components. By calculating reciprocals of the resistances, we find that:\[R_{eq} = \frac{1}{1.25} = 0.80\, \Omega\]This value, \( 0.80 \, \Omega \), is smaller than the smallest individual resistor in the parallel setup. Why is that important? In a parallel circuit, having multiple paths for current leads to lower overall resistance, allowing more current to flow.

Electrical power quantifies the rate at which energy is used or transferred in a circuit. It's calculated as the product of current and voltage but can also be found using resistance and current:\[ P = I^2R \]In this exercise:

• The power dissipated in the 1.60 \( \Omega \) resistor is \( 490 \, W \), as it carries the highest current.
• The other resistors dissipate \( 327 \, W \) and \( 163 \, W \) respectively.

These calculations help us see how resistors convert electrical energy into heat. Notably, even with different resistances, the resistor with the lowest resistance dissipates the most power because of higher current (from Ohm’s Law \( I = \frac{V}{R} \)).
In simple terms, less resistance in a parallel circuit allows more current, leading to more energy (power) converted to heat.

Determining the current through each resistor in a parallel circuit is straightforward using Ohm’s Law without needing complex math:
1. Calculate each current using \( I = \frac{V}{R} \) with \( V = 28.0 \, V \). For each resistor in this exercise:

• \( I_1 = \frac{28.0}{1.60} = 17.5 \, A \)
• \( I_2 = \frac{28.0}{2.40} = 11.67 \, A \)
• \( I_3 = \frac{28.0}{4.80} = 5.83 \, A \)

2. Total current through the battery is the sum of these currents: \[ I_{total} = I_1 + I_2 + I_3 = 35 \, A \]Understanding these steps shows that even in real-world applications, current divides according to the resistance across each path in a parallel circuit. This division allows for varying current amounts, depending on each pathway's resistance.