Objective: Calculate the current flowing through the 20 \(\Omega\) resistor in the given circuit.
The circuit consists of:
First, find the equivalent resistance of the parallel resistors (\(R_p\)):
\( R_p = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{20\ \Omega \times 30\ \Omega}{20\ \Omega + 30\ \Omega} = \frac{600\ \Omega^2}{50\ \Omega} = 12\ \Omega \)
Next, find the total equivalent resistance (\(R_{total}\)) of the circuit, considering \(R_p\) is in series with \(R_3\):
\( R_{total} = R_p + R_3 = 12\ \Omega + 8\ \Omega = 20\ \Omega \)
Calculate the total current (\(I_{total}\)) flowing from the battery using Ohm's Law (\(V = I \times R\)):
\( I_{total} = \frac{V_{battery}}{R_{total}} = \frac{12\ \text{V}}{20\ \Omega} = 0.6\ \text{A} \)
This total current (\(I_{total}\)) flows through the series resistor (\(R_3\)) and then splits between the parallel resistors (\(R_1\) and \(R_2\)). The voltage drop across the parallel section (\(V_p\)) is:
\( V_p = I_{total} \times R_p = 0.6\ \text{A} \times 12\ \Omega = 7.2\ \text{V} \)
The voltage across the parallel combination (\(V_p\)) is the same for both the \(20\ \Omega\) and \(30\ \Omega\) resistors. Apply Ohm's Law to find the current (\(I_{20\ \Omega}\)) through the \(20\ \Omega\) resistor (\(R_1\)):
\( I_{20\ \Omega} = \frac{V_p}{R_1} = \frac{7.2\ \text{V}}{20\ \Omega} = 0.36\ \text{A} \)
Therefore, the current through the \(20\ \Omega\) resistor is 0.36 A.
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