Net Current in DC Circuit with Capacitor at Steady State ⚡ | JEE Mains Physics Circuit Analysis PYQ

Master the concept of a capacitor in a DC circuit at steady state with this solved JEE Mains Physics PYQ! 💡 The question asks for the net current flowing in the circuit shown, which includes resistors and a single capacitor, connected to a DC voltage source.Key Physics Principle (Steady State):When a DC voltage source is connected to a circuit containing a capacitor, after a long time (at steady state), the capacitor is fully charged. In this state, the capacitor acts as an open circuit (infinite resistance) in its branch, meaning no current flows through the capacitor branch.Circuit Analysis and Key Steps Covered in the Video:Identify the Steady State Condition: The net current in a DC circuit is calculated after the capacitor has fully charged, meaning the circuit is in a steady state.Open Circuit the Capacitor Branch: Since the capacitor ($1 \mu \text{F}$) acts as an open circuit, the entire branch containing the $2.5 \Omega$ resistor, the $1 \mu \text{F}$ capacitor, and the $1 \Omega$ resistor is effectively broken. No current flows through this path.Redraw the Simplified Circuit: The simplified circuit consists only of the resistors parallel to the voltage source, excluding the capacitor branch.Left Section: Resistors $2\Omega$ and $4\Omega$ are in parallel.Middle Section: Resistors $3\Omega$ and $6\Omega$ are in parallel.Right Section: Resistors $5\Omega$ and $4\Omega$ are in parallel (The $8\Omega$ resistor is part of the broken capacitor branch and is ignored).The Actual Effective Circuit:The $2\Omega$ and $4\Omega$ are in parallel.The $3\Omega$ and $6\Omega$ are in parallel.The $5\Omega$ and $4\Omega$ are in parallel.However, observe the main structure: the only path through which current flows from the positive terminal to the negative terminal (other than the now-open capacitor branch) is the parallel combination of all resistors directly connected across the $2 \text{ V}$ battery.Simplify for Net Current: The circuit consists of multiple parallel branches across the $2 \text{ V}$ source.Branch 1: $2 \Omega$Branch 2: $4 \Omega$Branch 3: $3 \Omega$Branch 4: $6 \Omega$Branch 5: $5 \Omega$Branch 6: $4 \Omega$ (The $1\Omega$ and $8\Omega$ are part of the open circuit path and ignored.)Calculate Total Resistance ($R_{eq}$):$$\frac{1}{R_{eq}} = \frac{1}{2} + \frac{1}{4} + \frac{1}{3} + \frac{1}{6} + \frac{1}{5} + \frac{1}{4}$$$$\frac{1}{R_{eq}} = (\frac{1}{2} + \frac{1}{4} + \frac{1}{4}) + (\frac{1}{3} + \frac{1}{6}) + \frac{1}{5}$$$$\frac{1}{R_{eq}} = (1) + (\frac{3}{6}) + \frac{1}{5} = 1 + \frac{1}{2} + \frac{1}{5} = \frac{10+5+2}{10} = \frac{17}{10}$$$$R_{eq} = \frac{10}{17} \Omega$$Calculate Net Current ($I$): Using Ohm's Law, $I = V/R_{eq}$.$$I = \frac{2 \text{ V}}{10/17 \Omega} = 2 \times \frac{17}{10} = \frac{34}{10} = \mathbf{3.4 \text{ A}}$$This problem is crucial for JEE Mains 2026 and JEE Mains 2027 preparation, solidifying your knowledge of DC circuits and Capacitors.👉 Subscribe to Solved Insights for more complex circuit problems solved simply!

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