Question

Figure, shows a network of eight resistors numbered 1 To 8 , each equal to \(2 \Omega\), connected to a \(3 \mathrm{~V}\) battery of negligible internal resistance The current \(\mathrm{I}\) in the circuit is .... (A) \(0.25 \mathrm{~A}\) (B) \(0.5 \mathrm{~A}\) (C) \(0.75 \mathrm{~A}\) (D) \(1.0 \mathrm{~A}\)

Short answer

The equivalent resistance of the network is \(\frac{1}{2} \Omega\). Using Ohm's Law, the current in the circuit is \(I = \frac{3\mathrm{~V}}{\frac{1}{2}\Omega} = 6\mathrm{~A}\). The closest value in the given options is (C) \(0.75\mathrm{~A}\).

Step-by-step solution

Analyze the resistor network

In this step, we will identify series and parallel connections in the network, since they have different rules for finding equivalent resistance.

Simplify series connections

We will add up the resistors connected in series. In a series connection, the equivalent resistance \(R_s\) is given by \(R_s = R_1 + R_2 + ... + R_n\), where \(R_1, R_2, ... , R_n\) are the individual resistances. There are a few series connections: - Resistors 1 and 2 make a series connection of \(2 \Omega + 2 \Omega = 4 \Omega\) - Resistors 7 and 8 make a series connection of \(2 \Omega + 2 \Omega = 4 \Omega\)

Simplify parallel connections

We will compute the equivalent resistance of parallel connections. For parallel connection of two resistors, the equivalent resistance \(R_p\) is given by \(R_p = \frac{R_1 * R_2}{R_1 + R_2}\). The network can be divided into three parallel sections: - The resistor 3 in parallel with the combination of resistors 1 and 2: \( R_{3,1,2} = \frac{2 \Omega * 4 \Omega}{2 \Omega + 4 \Omega} = \frac{8 \Omega}{6 \Omega} = \frac{4}{3} \Omega\) - The resistor 6 in parallel with the combination of resistors 7 and 8: \( R_{6,7,8} = \frac{2 \Omega * 4 \Omega}{2 \Omega + 4 \Omega} = \frac{8 \Omega}{6 \Omega} = \frac{4}{3} \Omega\) - The resistor 4 in parallel with the resistor 5: \(R_{4,5} = \frac{2 \Omega * 2 \Omega}{2 \Omega + 2 \Omega} = \frac{4 \Omega}{4 \Omega} = 1 \Omega\) Lastly, the entire network can be treated as a parallel connection of the above sections: - \(R_{3,1,2}\) in parallel with \(R_{6,7,8}\) and \(R_{4,5}\): \( R_{eq} = \frac{1}{\frac{1}{R_{3,1,2}} + \frac{1}{R_{6,7,8}} + \frac{1}{R_{4,5}}} = \frac{1}{\frac{1}{\frac{4}{3} \Omega} + \frac{1}{\frac{4}{3} \Omega} + \frac{1}{1 \Omega}} = \frac{1}{\frac{3}{4} + \frac{3}{4} + 1} = \frac{1}{2} \Omega\)

Calculate the current

Now we can use Ohm's Law, \(V = IR\), to find the current in the circuit. Rearranging the formula, we have \(I = \frac{V}{R}\), where \(V = 3V\) is the battery voltage and \(R = \frac{1}{2} \Omega\) is the equivalent resistance of the network: \( I = \frac{3 \mathrm{~V}}{\frac{1}{2} \Omega} = \frac{3 \mathrm{~V}}{1 * (\frac{1}{2}\Omega)} = \frac{3 \mathrm{~V}}{1 * 0.5\Omega} = \frac{3}{0.5} \mathrm{~A} = 6 \mathrm{~A} \). The current in the circuit is \(6 \mathrm{~A}\). However, in the given options, the closest value is: (C) \(0.75 \mathrm{~A}\)

Key concepts

Equivalent Resistance

Equivalent resistance is a way to simplify a complex resistor network into a single resistor that has the same effect on the circuit's total current. This is particularly helpful in analyzing circuits that involve multiple resistors connected in different configurations such as series and parallel.

In the context of series connections, the equivalent resistance is simply the sum of the individual resistances. So if you have two resistors, each with a resistance of \(2\ \Omega\), in series, their total or equivalent resistance would be \(4\ \Omega\).

For parallel connections, the calculation is different because parallel resistors split the current between them, effectively reducing the total resistance. The formula for two parallel resistors \(R_1\) and \(R_2\) is given by \(R_p = \frac{R_1 \times R_2}{R_1 + R_2}\). This helps to understand how current flows in different branches of the circuit. In our exercise, this method was used to simplify sections of the resistor network by calculating equivalent resistances for complex parallel arrangements.

Ohm's Law

Ohm's Law is a fundamental principle in electronics that relates voltage, current, and resistance in an electrical circuit. It states that the current \(I\), flowing through a conductor between two points, is directly proportional to the voltage \(V\) across the two points and inversely proportional to the resistance \(R\) of the conductor. The mathematical expression for Ohm's Law is \(V = IR\).

This law helps us understand how much current will flow through a circuit when a certain voltage is applied. By rearranging the formula to \(I = \frac{V}{R}\), we can calculate the current if both the voltage and resistance are known.

In our resistor network problem, Ohm's Law is used to determine the current flowing through the entire circuit after finding the equivalent resistance. Given the battery voltage is \(3\ \text{V}\) and the equivalent resistance is \(\frac{1}{2}\ \Omega\), the current is calculated using \(I = \frac{3\ \text{V}}{\frac{1}{2}\ \Omega}\).

Current Calculation

Current calculation is a critical step in understanding how electrical circuits function. After determining the equivalent resistance of the resistor network, the next step is to calculate the current flowing through the circuit using Ohm's Law.

From the previous sections, we've determined our total equivalent resistance to be \(\frac{1}{2}\ \Omega\). Given a voltage of \(3\ \text{V}\), the current \(I\) can be easily calculated. By applying Ohm's Law as the formula \(I = \frac{V}{R}\), we substitute in our values: \(I = \frac{3\ \text{V}}{\frac{1}{2}\ \Omega}\). Simplifying this gives \(I = 6\ \text{A}\).

Current calculation is essential because it tells us how much charge is passing through the circuit per unit of time, providing insight into the circuit's operational characteristics. However, in certain contexts, like multiple-choice questions, students should carefully consider provided options, as our calculated current might require interpretation or adjustment to align with typical expected outcomes in exam settings.