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Question #42698

Difficulty: 8/10 | 🟢 Free

The Question

A semi-infinite ladder network is constructed with alternating series and shunt branches. The input terminals are at the left end. For all integers $n \ge 1$ , the $n$ -th series branch contains a capacitor of capacitance $C$ (in Farads) and a resistor of resistance $R_n$ (in Ohms) connected in series. The $n$ -th shunt branch contains a capacitor of capacitance $2C$ (in Farads) and a resistor of resistance $r_n$ (in Ohms) connected in series. The resistance values $R_n$ and $r_n$ are arbitrary, strictly positive finite constants. At $t=0$ , an ideal DC voltage source of EMF $V_0$ (in Volts) is connected across the input terminals. Assuming all capacitors are initially uncharged, what is the total energy dissipated as heat (in Joules) in all the resistors of the network as $t \to \infty$ ?

Options

A.
$\frac{\sqrt{3}-1}{2} C V_0^2$
B.
$\frac{\sqrt{5}-1}{4} C V_0^2$
C.
$(\sqrt{3}-1) C V_0^2$
D.
$\frac{\sqrt{3}+1}{2} C V_0^2$

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Intuition_Breaker

Train your brain to resist "obvious" answers that are designed to deceive your intuition.

Concepts Behind This Question

Infinite Series Capacitance|Resistivity vs Resistance|Integration

Why This Question Is Tricky

Common Mistakes

This question tests the application of macroscopic energy conservation in complex transient circuits, specifically targeting the misconception that Joule heating ( $\int I^2 R dt$ ) must explicitly depend on the individual resistance values.

Students often fall into the trap of attempting to set up infinite systems of differential equations for the transient currents, leading to a 'Lethal Collision' of mathematical complexity.

The core physical principle is that for any linear capacitive network charged by a constant DC source, the total heat dissipated is exactly equal to the final energy stored ( $\frac{1}{2} C_{eq} V_0^2$ ), regardless of the intermediate resistive paths, provided steady state is reached.

Distractor options anticipate errors in calculating the equivalent capacitance of the infinite ladder (such as confusing it with the standard $C$ and $C$ golden ratio ladder) or forgetting the factor of $1/2$ when applying the work-energy theorem.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Electrostatic Potential and Capacitance - Class 12
Concept	Infinite Series Capacitance
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free