Step 1: Understanding the Concept:
To find the heat dissipated in a specific resistor, we must first determine the voltage drop across it or the current flowing through it. Nodal analysis is the most efficient method to find the potentials in a multi-branch parallel circuit.
Step 2: Key Formula or Approach:
Kirchhoff's Current Law (Nodal Analysis): \(\sum I = 0\) at any node.
Joule's Heating Law: \(H = \frac{V^2}{R} t\)
Step 3: Detailed Explanation:
Let the common node on the right be at \(0 \, \text{V}\) and the common node on the left be at \(x \, \text{V}\).
Applying KCL at node \(x\):
\[ \frac{x - 3}{6} + \frac{x - 0}{4} + \frac{x - 2}{2} = 0 \]
Multiply the entire equation by the LCM (12) to clear denominators:
\[ 2(x - 3) + 3x + 6(x - 2) = 0 \]
\[ 2x - 6 + 3x + 6x - 12 = 0 \]
\[ 11x - 18 = 0 \implies x = \frac{18}{11} \, \text{V} \]
The voltage drop across the \(6 \, \Omega\) resistor is \(V_{6\Omega} = x - 3\):
\[ V_{6\Omega} = \frac{18}{11} - 3 = \frac{18 - 33}{11} = -\frac{15}{11} \, \text{V} \]
The magnitude of the voltage is \(\frac{15}{11} \, \text{V}\).
Now calculate the heat dissipated in \(t = 100 \, \text{s}\):
\[ H = \frac{V_{6\Omega}^2}{R} \times t = \frac{\left(\frac{15}{11}\right)^2}{6} \times 100 \]
\[ H = \frac{225}{121 \times 6} \times 100 = \frac{22500}{726} \approx 30.99 \, \text{J} \]
Rounding to the nearest integer gives \(31 \, \text{J}\).
Step 4: Final Answer:
The heat dissipated in the \(6\Omega\) resistor is \(31 \, \text{J}\).
