Short Answer 
The total heat dissipated in resistors due to the discharge of a capacitor is calculated by deriving individual heat values from each resistor using an infinite geometric series. This results in the formula H = Q₀²/6C, confirming the outcome corresponds to option (b).
Step 1: Understanding Heat Dissipation in Resistors
The total energy dissipated as heat in a single resistor is formulated as Q¬≤/2C, where Q represents the charge stored in the capacitor and C is the capacitance. In this scenario, as the capacitor discharges through an infinite configuration of resistors, the initial charge Q‚ÇÄ becomes distributed among the resistors in a geometric progression. Each resistor’s heat dissipation will contribute to the overall heat loss in the system.
Step 2: Deriving Charge and Heat for Each Resistor
The charge carried by the nth resistor (Qₙ) is calculated using the formula: Q₀ * (1/2)^(n-1). Consequently, the heat dissipated in the nth resistor (Hₙ) is assessed with the equation: Qₙ²/2C. By substituting the value of Qₙ in this formula, we can express Hₙ as: (Q₀²/2C) * (1/4)^(n-1). This shows how the heat reduces with each subsequent resistor in the ladder.
Step 3: Calculating Total Heat Dissipation
To find the total heat dissipated through all horizontal resistors, we sum the heat from each resistor from n=1 to infinity. Utilizing the sum of an infinite geometric series, we derive the total heat as: H = (Q₀²/2C) * (4/3). After simplification, this results in the final expression: H = Q₀²/6C, which confirms that the result corresponds to option (b) of the problem.