Question

The power dissipated across the \(16\Omega\) resistor in the circuit is \(2\) watts. The power dissipated in watt units across the \(4\Omega\) resistor is:

• A. \(2.38\,\text{W}\)
• B. \(0.64\,\text{W}\)
• C. \(1.28\,\text{W}\)
• D. \(4.28\,\text{W}\)

Hint:

In parallel branches, voltage remains the same. In series resistors, current remains the same. Use these facts carefully to calculate power.

Answer 1

The Correct Option is C

Step 1: Identify the circuit arrangement.
The \(16\Omega\) resistor is connected in one branch, while the \(6\Omega\) and \(4\Omega\) resistors are connected in series in the other branch.

Step 2: Understand voltage across parallel branches.
Since both branches are connected in parallel, the potential difference across the \(16\Omega\) branch and the \((6\Omega + 4\Omega)\) branch is the same.

Step 3: Use power formula for the \(16\Omega\) resistor.
\[ P = \frac{V^2}{R} \]
Given,
\[ P = 2\,\text{W} \]
\[ R = 16\Omega \]
\[ 2 = \frac{V^2}{16} \]

Step 4: Find voltage across the parallel network.
\[ V^2 = 32 \]
\[ V = \sqrt{32} \]

Step 5: Find resistance of the lower branch.
The \(6\Omega\) and \(4\Omega\) resistors are in series, so:
\[ R_{\text{lower}} = 6 + 4 = 10\Omega \]

Step 6: Find current through the lower branch.
\[ I = \frac{V}{R_{\text{lower}}} \]
\[ I^2 = \frac{V^2}{R_{\text{lower}}^2} \]
\[ I^2 = \frac{32}{10^2} \]
\[ I^2 = \frac{32}{100} \]

Step 7: Find power across the \(4\Omega\) resistor.
Since \(6\Omega\) and \(4\Omega\) are in series, the same current flows through the \(4\Omega\) resistor.
\[ P_4 = I^2R \]
\[ P_4 = \frac{32}{100} \times 4 \]
\[ P_4 = \frac{128}{100} \]
\[ P_4 = 1.28\,\text{W} \]
\[ \boxed{1.28\,\text{W}} \]