This problem requires us to find the current flowing through a specific resistor in a circuit that combines both series and parallel connections. To solve this, we need to simplify the circuit step-by-step by calculating the equivalent resistance of the parallel part, then the total resistance of the entire circuit, and finally use Ohm's Law and the principle of current division (or voltage across parallel components) to find the current in the desired resistor.

Understanding the Circuit Configuration

The circuit consists of:

• Two resistors (12 \(\Omega\) and 24 \(\Omega\)) connected in parallel.
• This parallel combination is connected in series with a 22 \(\Omega\) resistor.
• The entire arrangement is connected across a 12 V battery.

Calculating Equivalent Resistance of Parallel Resistors

When resistors are connected in parallel, the reciprocal of the equivalent resistance (\(R_p\)) is the sum of the reciprocals of individual resistances. For the 12 \(\Omega\) (\(R_1\)) and 24 \(\Omega\) (\(R_2\)) resistors in parallel:

The formula is: \(\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2}\)

Substituting the values:

\(\frac{1}{R_p} = \frac{1}{12 \, \Omega} + \frac{1}{24 \, \Omega}\)

To add these fractions, we find a common denominator, which is 24:

\(\frac{1}{R_p} = \frac{2}{24 \, \Omega} + \frac{1}{24 \, \Omega} = \frac{2 + 1}{24 \, \Omega} = \frac{3}{24 \, \Omega}\)

Now, we find \(R_p\) by taking the reciprocal:

\(R_p = \frac{24}{3} \, \Omega = 8 \, \Omega\)

So, the equivalent resistance of the parallel combination is 8 \(\Omega\).

Calculating Total Circuit Resistance

The 8 \(\Omega\) equivalent resistance of the parallel section is in series with the 22 \(\Omega\) resistor (\(R_s\)). When resistors are in series, the total resistance (\(R_{total}\)) is simply the sum of the individual resistances:

\(R_{total} = R_p + R_s\)

\(R_{total} = 8 \, \Omega + 22 \, \Omega = 30 \, \Omega\)

The total resistance of the entire circuit is 30 \(\Omega\).

Calculating Total Current from the Battery

We can use Ohm's Law (\(V = IR\)) to find the total current (\(I_{total}\)) drawn from the 12 V battery (\(V_{total}\)). Rearranging the formula, \(I = V/R\).

\(I_{total} = \frac{V_{total}}{R_{total}}\)

\(I_{total} = \frac{12 \, V}{30 \, \Omega} = \frac{12}{30} \, A\)

Simplifying the fraction:

\(I_{total} = \frac{2}{5} \, A\)

This total current flows through the 22 \(\Omega\) series resistor and then splits between the 12 \(\Omega\) and 24 \(\Omega\) resistors in the parallel branch.

Calculating Current in the 24 \(\Omega\) Resistor

The total current (\(I_{total} = \frac{2}{5} \, A\)) enters the parallel branch containing the 12 \(\Omega\) and 24 \(\Omega\) resistors. We need to find the current through the 24 \(\Omega\) resistor (\(I_{24\Omega}\)).

We can use the current division rule. For two resistors \(R_1\) and \(R_2\) in parallel, the current through \(R_2\) (\(I_2\)) is given by:

\(I_2 = I_{total} \times \frac{R_1}{R_1 + R_2}\)

In our case, \(R_1 = 12 \, \Omega\), \(R_2 = 24 \, \Omega\), and \(I_{total} = \frac{2}{5} \, A\). We want to find the current through the 24 \(\Omega\) resistor (which is \(R_2\) in the formula, so \(R_1\) in the numerator refers to the other resistor, 12 \(\Omega\)).

\(I_{24\Omega} = I_{total} \times \frac{R_{12\Omega}}{R_{12\Omega} + R_{24\Omega}}\)

\(I_{24\Omega} = \frac{2}{5} \, A \times \frac{12 \, \Omega}{12 \, \Omega + 24 \, \Omega}\)

\(I_{24\Omega} = \frac{2}{5} \, A \times \frac{12 \, \Omega}{36 \, \Omega}\)

\(I_{24\Omega} = \frac{2}{5} \, A \times \frac{1}{3}\)

\(I_{24\Omega} = \frac{2 \times 1}{5 \times 3} \, A = \frac{2}{15} \, A\)

Alternatively, we could find the voltage across the parallel branch (\(V_{parallel}\)) using Ohm's Law (\(V = IR\)). Since the total current flows through the equivalent parallel resistance (\(R_p\)):

\(V_{parallel} = I_{total} \times R_p\)

\(V_{parallel} = \frac{2}{5} \, A \times 8 \, \Omega = \frac{16}{5} \, V\)

The voltage across each resistor in a parallel combination is the same. So, the voltage across the 24 \(\Omega\) resistor is \(\frac{16}{5} \, V\). We can use Ohm's Law again (\(I = V/R\)) to find the current through the 24 \(\Omega\) resistor:

\(I_{24\Omega} = \frac{V_{parallel}}{R_{24\Omega}}\)

\(I_{24\Omega} = \frac{16/5 \, V}{24 \, \Omega} = \frac{16}{5 \times 24} \, A = \frac{16}{120} \, A\)

Simplifying the fraction:

\(I_{24\Omega} = \frac{8}{60} \, A = \frac{4}{30} \, A = \frac{2}{15} \, A\)

Both methods give the same result.

Summary of Steps and Results

The current in the 24 \(\Omega\) resistor is \(\frac{2}{15} \, A\).

Revision Table: Key Concepts in Circuit Analysis

Additional Information: Analyzing Series Parallel Circuits

Analyzing circuits with both series and parallel components is a fundamental skill in electricity. The key is to simplify the circuit in stages. Start by finding the equivalent resistance of the smallest parallel or series groups. Replace these groups with their equivalent resistances and redraw the circuit if necessary. Continue this process until you have a simple series or parallel circuit across the voltage source. Once the total equivalent resistance is found, you can calculate the total current from the source using Ohm's Law.

After finding the total current, you can work backward to find the current and voltage across individual components:

• For resistors in series, the current through each resistor is the same as the total current flowing into that series section. The voltage across each resistor can be found using Ohm's Law (\(V=IR\)).
• For resistors in parallel, the voltage across each resistor is the same as the voltage across the parallel combination. This voltage can be found by multiplying the total current entering the parallel branch by the equivalent resistance of the parallel branch (\(V_{parallel} = I_{total\_branch} \times R_{parallel\_eq}\)). Once the voltage is known, the current through each parallel resistor can be found using Ohm's Law (\(I = V_{parallel} / R\)). Alternatively, current division can be used directly if the resistors and total branch current are known.

Mastering these steps allows you to solve complex series-parallel circuit problems.