A $4\text{ A}$ current source feeds $8 \ \Omega$ and $4 \ \Omega$ in parallel. Power in $8 \ \Omega$ is:
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An alternative approach is to find the equivalent resistance \(R_{eq}\) first:
\[ R_{eq} = \frac{8 \times 4}{8 + 4} = \frac{32}{12} = \frac{8}{3}\ \Omega \]
The voltage across the parallel combination is:
\[ V = I_{\text{total}} \times R_{eq} = 4 \times \frac{8}{3} = \frac{32}{3}\text{ V} \]
The power in the \(8\ \Omega\) resistor is:
\[ P = \frac{V^2}{R_1} = \frac{(32/3)^2}{8} = \frac{1024}{9 \times 8} = \frac{128}{9} \approx 14.22\text{ W} \]
This cross-verification helps avoid calculation errors.
Step 1: Understanding the Question:
The question asks for the electrical power dissipated in an \(8\ \Omega\) resistor which is connected in parallel with a \(4\ \Omega\) resistor, both being fed by a constant \(4\text{ A}\) current source. Step 2: Key Formula or Approach:
1. Use the current division rule to find the current flowing through the \(8\ \Omega\) resistor:
\[ I_1 = I_{\text{total}} \times \frac{R_2}{R_1 + R_2} \]
where \(R_1 = 8\ \Omega\), \(R_2 = 4\ \Omega\), and \(I_{\text{total}} = 4\text{ A}\).
2. Calculate the power dissipated in the \(8\ \Omega\) resistor using:
\[ P = I_1^2 \cdot R_1 \]
Step 3: Detailed Explanation:
• Apply the current division rule to determine the current \(I_1\) in the \(8\ \Omega\) branch:
\[ I_1 = 4\text{ A} \times \frac{4\ \Omega}{8\ \Omega + 4\ \Omega} = 4 \times \frac{4}{12} = \frac{4}{3}\text{ A} \approx 1.333\text{ A} \]
• Calculate the power \(P_1\) consumed by the \(8\ \Omega\) resistor:
\[ P_1 = I_1^2 \times R_1 \]
\[ P_1 = \left(\frac{4}{3}\right)^2 \times 8 \]
\[ P_1 = \frac{16}{9} \times 8 \]
\[ P_1 = \frac{128}{9}\text{ W} \approx 14.22\text{ W} \]
Step 4: Final Answer:
The power dissipated in the \(8\ \Omega\) resistor is \(14.22\text{ W}\).
