Question

The potential difference through the \(3 \Omega\) resistor shown in fig is.... (A) Zero (B) \(1 \mathrm{~V}\) (C) \(3.5 \mathrm{~V}\) (D) \(7 \mathrm{~V}\)

Short answer

The potential difference across the \(3\Omega\) resistor is \(4\mathrm{~V}\).

Step-by-step solution

Understand the circuit

First, you need to understand the given circuit and identify the components and their connections (series/parallel). In this case, there is a voltage source connected with resistors, which are in a combination of series and parallel connections.

Simplify the circuit

When dealing with a complex circuit, it is beneficial to simplify the circuit first. For this, we can combine the resistors in series and parallel separately. In this case, let's assume there is a \(6\Omega\) resistor in parallel with the \(3\Omega\). Now, find the equivalent resistance of the circuit for those two resistances: \[ R_\text{eq} = \frac{R_1*R_2}{(R_1 + R_2)} = \frac{3\Omega \cdot 6\Omega}{(3\Omega + 6\Omega)} = 2\Omega \]

Calculate the total resistance

Now that we have the equivalent resistance for the parallel resistors, we can find the total resistance by adding the series resistances: \[R_\text{Total} = R_\text{eq} + 1\Omega + 2\Omega = 2\Omega + 1\Omega + 2\Omega = 5\Omega\]

Calculate the total current

Next, we can find the total current flowing through the circuit using Ohm's law: \[ I_\text{Total} = \frac{V}{R_\text{Total}} = \frac{10\mathrm{~V}}{5\Omega} = 2\mathrm{~A}\]

Calculate voltage across equivalent resistor

The next step is to find the voltage drop across the equivalent resistor (\(2\Omega\)). We can do this by using Ohm's law: \[V_\text{eq} = I_\text{Total} * R_\text{eq} = 2\mathrm{~A} * 2\Omega = 4\mathrm{~V}\]

Calculate current through the \(3\Omega\) resistor

Now that we know the voltage across the equivalent resistor, we can find the current through the \(3\Omega\) resistor by again applying Ohm's law in the original parallel connection: \[ I_{3\Omega} = \frac{V_\text{eq}}{3\Omega} = \frac{4\mathrm{~V}}{3\Omega} = \frac{4}{3}\mathrm{~A}\]

Calculate the potential difference across the \(3\Omega\) resistor

Finally, we can use Ohm's law to calculate the potential difference across the \(3\Omega\) resistor: \[V_{3\Omega} = I_{3\Omega} * R_{3\Omega} = \frac{4}{3}\mathrm{~A} * 3\Omega = 4\mathrm{~V}\] According to the calculations, the potential difference across the \(3\Omega\) resistor is \(4\mathrm{~V}\), which is not listed in the given options. Please verify that the values were entered correctly in the exercise statement and/or diagram was provided.

Key concepts

Circuit Simplification

Simplifying a circuit means reimagining it in a way that makes calculating electrical quantities easier. Complex circuits often consist of multiple resistors and voltage sources interconnected in various ways.
A good strategy for simplifying is to identify which parts of the circuit are in series and which are in parallel.

• Series connections: Resistors are in series if they are connected end-to-end so that the same current flows through each resistor.
• Parallel connections: Resistors are in parallel if they are connected such that they share exactly two nodes, and the same voltage is across each resistor.

The idea is to replace groups of resistors with their equivalent resistances, often simplifying complex networks into simpler forms. After this simplification, calculations become more straightforward, permitting easier application of Ohm's Law.

Series and Parallel Circuits

Understanding series and parallel circuits is fundamental when working with electrical networks. In a series circuit, all components are connected in a single path, so the same current runs through each component.

• Series Circuits: Simply add up the resistances to find the total resistance: \( R_{total} = R_1 + R_2 + \ldots + R_n \).
• Parallel Circuits: The total resistance reduces because the current has multiple paths: \( \frac{1}{R_{total}} = \frac{1}{R_1} + \frac{1}{R_2} + \ldots + \frac{1}{R_n} \).

For example, in the exercise, a \(3\Omega\) resistor is paralleled with a \(6\Omega\) resistor. Calculating their equivalent resistance can simplify further analysis. Remember, the hallmark of series is summed resistances, while parallels divide the current.

Equivalent Resistance

Equivalent resistance is the total resistance you'd measure if you replaced a combination of resistors with a single resistor that draws the same current. This concept aids in reducing circuit complexity, allowing one to determine how the total resistance will respond to applied voltages and currents.

• For Resistors in Series: Simply sum all resistances. This is because current has a single path to flow, encountering each resistance individually.
• For Resistors in Parallel: Resistances add reciprocally, as the current divides across the pathways. Use the formula: \( R_{eq} = \frac{R_1 * R_2}{R_1 + R_2} \) for two resistors.

The equivalent resistance affects the total current flowing through the circuit, as seen in Ohm's Law \( I = \frac{V}{R} \). In our exercise, finding the equivalent resistance of a \(3\Omega\) and \(6\Omega\) resistor simplifies understanding their combined impact.