Capacitors & Resistance: JEE Mains

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Capacitors: Basics and Combinations Capacitance: $C = \frac{Q}{V}$ (Unit: Farad, F) Energy Stored: $U = \frac{1}{2}CV^2 = \frac{1}{2}\frac{Q^2}{C} = \frac{1}{2}QV$ Capacitance of Parallel Plate Capacitor: $C = \frac{\epsilon A}{d}$ With dielectric: $C_k = kC_0 = \frac{k\epsilon_0 A}{d}$ Series Combination: $\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \dots$ Charge $Q$ is same across each capacitor. Voltage divides: $V = V_1 + V_2 + \dots$ Parallel Combination: $C_{eq} = C_1 + C_2 + \dots$ Voltage $V$ is same across each capacitor. Charge divides: $Q = Q_1 + Q_2 + \dots$ Resistors: Basics and Combinations Ohm's Law: $V = IR$ (Unit: Ohm, $\Omega$) Resistance: $R = \rho \frac{L}{A}$ $\rho$: resistivity, $L$: length, $A$: cross-sectional area. Series Combination: $R_{eq} = R_1 + R_2 + \dots$ Current $I$ is same through each resistor. Voltage divides: $V = V_1 + V_2 + \dots$ Parallel Combination: $\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \dots$ Voltage $V$ is same across each resistor. Current divides: $I = I_1 + I_2 + \dots$ Electrical Power: $P = VI = I^2 R = \frac{V^2}{R}$ (Unit: Watt, W) RC Circuits (Charging/Discharging) Charging of a Capacitor: Charge: $Q(t) = Q_0 (1 - e^{-t/\tau})$ Current: $I(t) = I_0 e^{-t/\tau}$ Voltage across capacitor: $V_C(t) = V_0 (1 - e^{-t/\tau})$ Discharging of a Capacitor: Charge: $Q(t) = Q_0 e^{-t/\tau}$ Current: $I(t) = -\frac{Q_0}{RC} e^{-t/\tau}$ Voltage across capacitor: $V_C(t) = V_0 e^{-t/\tau}$ Time Constant: $\tau = RC$ (Unit: seconds, s) Common Question Types for JEE Mains 1. Equivalent Capacitance/Resistance Simple Series/Parallel: Direct application of formulas. Complex Networks: Wheatstone Bridge: If balanced ($\frac{C_1}{C_2} = \frac{C_3}{C_4}$ or $\frac{R_1}{R_2} = \frac{R_3}{R_4}$), the middle branch can be removed. Symmetry: Identify lines of symmetry to simplify circuits. Infinite Ladder: Assume $R_{eq}$ (or $C_{eq}$) for the entire ladder is the same as for a section, then solve. Delta-Star/Star-Delta Transformation: For complex 3-terminal networks (less common for basic JEE, but good to know). 2. Charge, Voltage, Current Distribution Kirchhoff's Laws: Junction Rule (KCL): $\sum I_{in} = \sum I_{out}$ Loop Rule (KVL): $\sum \Delta V = 0$ in any closed loop. Steady State of RC Circuits: After a long time ($t \to \infty$), capacitor acts as an open circuit (no current flows through it). Trick: Remove the capacitor from the circuit and solve for voltages. Voltage across capacitor becomes constant. Just After Switching ($t=0^+$): A capacitor acts as a short circuit (zero resistance) if uncharged. Trick: Replace the uncharged capacitor with a wire and solve for currents. A capacitor acts as a voltage source if initially charged. 3. Energy Stored and Dissipated Energy Stored in Capacitor: $U = \frac{1}{2}CV^2$ Energy Dissipated in Resistor: $E_{dissipated} = \int P dt = \int I^2 R dt$ Energy Loss in Capacitor Redesign/Sharing: When capacitors are connected or reconnected, charge redistribution occurs, leading to energy loss, usually in the form of heat. Common potential: $V_{common} = \frac{Q_{total}}{C_{total}}$ Energy loss: $\Delta U = U_{initial} - U_{final}$ Trick: For two capacitors $C_1, C_2$ with initial voltages $V_1, V_2$ connected in parallel, energy loss is $\frac{1}{2} \frac{C_1 C_2}{C_1+C_2} (V_1-V_2)^2$. 4. Dielectrics in Capacitors Introducing Dielectric: Battery connected (V constant): $C \to kC$, $Q \to kQ$, $U \to kU$. Battery disconnected (Q constant): $C \to kC$, $V \to V/k$, $U \to U/k$. Partially Filled Capacitor: Series combination: $C = \frac{\epsilon_0 A}{\left(d-t+\frac{t}{k}\right)}$ Parallel combination: $C = \frac{\epsilon_0 A}{d} \left( \frac{k_1 A_1 + k_2 A_2}{A} \right)$ (if divided parallel to plates) 5. Galvanometer, Ammeter, Voltmeter Galvanometer: Basic current-sensing device. Ammeter: Low resistance shunt resistor connected in parallel with galvanometer. $I_g R_g = (I - I_g) S \implies S = \frac{I_g R_g}{I - I_g}$ Voltmeter: High resistance resistor connected in series with galvanometer. $V = I_g (R_g + R) \implies R = \frac{V}{I_g} - R_g$ 6. Potentiometer and Meter Bridge Meter Bridge: Based on Wheatstone bridge principle. $\frac{R_1}{R_2} = \frac{l_1}{l_2}$ (where $l_1+l_2=100$ cm) Potentiometer: Measures EMF/potential difference without drawing current. Principle: Potential drop across any length of wire is directly proportional to its length ($V \propto L$). Comparison of EMFs: $\frac{E_1}{E_2} = \frac{l_1}{l_2}$ Internal Resistance of Cell: $r = R \left(\frac{l_1}{l_2} - 1\right)$ Circuit Analysis Examples & Tricks Example 1: Charge Stored in a Capacitor (Steady State) Consider a circuit with a battery (V), resistors ($R_1, R_2, R_3$), and a capacitor (C) connected. Find the charge on the capacitor in steady state. Trick (Steady State): In steady state, the capacitor acts as an open circuit. No current flows through the branch containing the capacitor. Steps: Remove the capacitor branch for current calculation. Calculate the current flowing through the resistors. For example, if $R_1$ is in series with $R_2$ and $R_3$ is parallel to $R_2$, find $I_{total}$. Find the voltage across the points where the capacitor is connected. This voltage will be the voltage across the capacitor, $V_C$. Example: If C is across $R_2$, then $V_C = I_{R_2} \cdot R_2$. Calculate charge $Q = C V_C$. Example 2: Equivalent Resistance (Complex Network) Find the equivalent resistance between points A and B in a complex network (e.g., a cube of resistors). Trick (Symmetry): Look for lines or planes of symmetry. Points on a line/plane of symmetry that are equidistant from the terminals (A and B) will have the same potential. Steps: Identify points with the same potential due to symmetry. Redraw the circuit by combining these equipotential points. This often simplifies the network into series/parallel combinations. Solve the simplified circuit. Trick (Folding Symmetry): If the circuit can be folded along a line such that components overlap perfectly, then components on either side of the fold line (if they are symmetrical) can be considered in parallel or removed depending on the specific symmetry. Example 3: Current just after switching (t=0+) A switch is closed at $t=0$. Find the current through a resistor just after closing the switch. Trick (t=0+): An uncharged capacitor acts as a short circuit (a plain wire). Steps: Replace all uncharged capacitors with a wire. Replace all charged capacitors with a voltage source equal to their initial voltage. Solve the resulting purely resistive circuit using Ohm's law and Kirchhoff's laws. Example 4: Energy Loss when Capacitors are Connected Two capacitors $C_1$ and $C_2$ are charged to $V_1$ and $V_2$ respectively. They are then connected in parallel (positive to positive, negative to negative). Find the energy loss. Trick (Common Potential): The total charge is conserved. The final voltage across both capacitors will be the same. Steps: Calculate initial charges: $Q_1 = C_1 V_1$, $Q_2 = C_2 V_2$. Calculate total charge: $Q_{total} = Q_1 + Q_2$. Calculate equivalent capacitance: $C_{eq} = C_1 + C_2$. Calculate common potential: $V_{common} = \frac{Q_{total}}{C_{eq}}$. Calculate initial stored energy: $U_{initial} = \frac{1}{2}C_1 V_1^2 + \frac{1}{2}C_2 V_2^2$. Calculate final stored energy: $U_{final} = \frac{1}{2}C_{eq} V_{common}^2$. Energy loss: $\Delta U = U_{initial} - U_{final}$. Direct formula for loss: $\Delta U = \frac{1}{2} \frac{C_1 C_2}{C_1+C_2} (V_1-V_2)^2$. General Tricks for Circuit Problems Node Analysis: Assign potentials to nodes, set one node to 0V (ground), and use KCL at each node to form equations. This is very powerful for complex resistor networks. Mesh Analysis: Assign loop currents and apply KVL to each independent loop. Useful for finding currents in specific branches. Source Transformation: Convert voltage sources with series resistors to current sources with parallel resistors, and vice versa, to simplify circuits. Superposition Theorem: For linear circuits, the response (current/voltage) in any component due to multiple independent sources is the algebraic sum of responses due to each source acting alone (other sources turned off). Thevenin's/Norton's Theorem: Simplify a complex linear circuit to an equivalent voltage source (Thevenin) or current source (Norton) with an equivalent resistance. Useful for finding current/voltage through a specific load. Potential Method: Assign potential at different points. If two points are connected by a wire, they have the same potential. Helps in identifying parallel combinations. JEE Mains 2024 Relevant Questions (Illustrative Examples) & Solutions JEE Mains 2024 (Shift 1, Jan 27) - Equivalent Resistance Question: Find the equivalent resistance between points A and B in the following circuit, where all resistors are $R$. A B R R R R R R R (This is a common Wheatstone bridge type problem, often disguised. The diagram shows a bridge with a central resistor.) Solution: Let the nodes be A, C, D, E, B. (A, C, D top, A-C-D-E-B bottom, C-E middle vertical) A B C D E F R R R R R R R This is a balanced Wheatstone bridge if we consider A-C-E-B path and A-D-F-B path. The resistor between C and F forms the bridge's cross-arm. The ratio $\frac{R_{AC}}{R_{AD}} = \frac{R}{R} = 1$ and $\frac{R_{CE}}{R_{DF}} = \frac{R}{R} = 1$. Since the ratios are equal, the bridge is balanced. Thus, the potential at C is equal to the potential at F. No current flows through the resistor between C and F. Remove the central resistor (between C and F). The circuit simplifies to: Upper arm: $R$ (A-C) in series with $R$ (C-E) in series with $R$ (E-B) $\implies 3R$ Lower arm: $R$ (A-D) in series with $R$ (D-F) in series with $R$ (F-B) $\implies 3R$ These two $3R$ arms are in parallel. Equivalent resistance $R_{eq} = \frac{3R \times 3R}{3R + 3R} = \frac{9R^2}{6R} = \frac{3R}{2}$. JEE Mains 2024 (Shift 2, Jan 29) - RC Circuit (Steady State) Question: In the given circuit, the switch S is closed for a long time. Find the charge stored on the capacitor of capacitance C. (Assume $V=10V$, $R_1=2\Omega$, $R_2=3\Omega$, $R_3=5\Omega$, $C=1\mu F$). V $R_1$ $R_2$ $R_3$ C P Q S (Switch) Solution: At steady state, the capacitor acts as an open circuit. This means no current flows through the branch containing C ($P-Q$ branch). The circuit effectively becomes a series combination of $R_1$, $R_2$, and $R_3$ connected to the voltage source V. Equivalent resistance $R_{eq} = R_1 + R_2 + R_3 = 2\Omega + 3\Omega + 5\Omega = 10\Omega$. The total current flowing through the circuit is $I = \frac{V}{R_{eq}} = \frac{10V}{10\Omega} = 1A$. The voltage across the capacitor is the potential difference between points P and Q ($V_{PQ}$). Since no current flows through the capacitor branch, $V_{PQ}$ is the potential drop across $R_3$. $V_C = V_{PQ} = I \cdot R_3 = 1A \cdot 5\Omega = 5V$. The charge stored on the capacitor is $Q = C \cdot V_C = (1 \mu F) \cdot (5V) = 5 \mu C$. JEE Mains 2024 (Shift 1, Feb 1) - Potentiometer Question: A potentiometer wire of length 400 cm and resistance $8\Omega$ is connected in series with a battery of EMF 2V and internal resistance $2\Omega$. An unknown cell of EMF $E_x$ is balanced at 200 cm. Find $E_x$. Solution: Total resistance of the primary circuit: $R_{total} = R_{wire} + r_{battery} = 8\Omega + 2\Omega = 10\Omega$. Total current in the primary circuit: $I_{main} = \frac{E_{battery}}{R_{total}} = \frac{2V}{10\Omega} = 0.2A$. Potential drop across the potentiometer wire: $V_{wire} = I_{main} \cdot R_{wire} = 0.2A \cdot 8\Omega = 1.6V$. Potential gradient of the wire: $k = \frac{V_{wire}}{L_{wire}} = \frac{1.6V}{400 \text{ cm}} = \frac{1.6V}{4 \text{ m}} = 0.4 V/m$. The unknown cell $E_x$ is balanced at length $l = 200 \text{ cm} = 2 \text{ m}$. Therefore, $E_x = k \cdot l = (0.4 V/m) \cdot (2 \text{ m}) = 0.8V$. JEE Mains 2024 (Various Shifts) - Power Dissipation in Resistors Question: In the circuit shown, if $R_1 = 1\Omega$, $R_2 = 2\Omega$, $R_3 = 3\Omega$, and the voltage source is $V=12V$. Find the power dissipated in resistor $R_2$. V $R_1$ $R_2$ $R_3$ Solution: The resistors $R_1, R_2, R_3$ are in series. Equivalent resistance $R_{eq} = R_1 + R_2 + R_3 = 1\Omega + 2\Omega + 3\Omega = 6\Omega$. Current flowing through the circuit: $I = \frac{V}{R_{eq}} = \frac{12V}{6\Omega} = 2A$. Since it's a series circuit, the same current $I=2A$ flows through $R_2$. Power dissipated in $R_2$: $P_2 = I^2 R_2 = (2A)^2 \cdot (2\Omega) = 4 \cdot 2 = 8W$. JEE Mains 2024 (Various Shifts) - Capacitors with Dielectrics Question: A parallel plate capacitor has capacitance $C_0$ in air. A dielectric slab of dielectric constant $k=4$ is inserted to fill the entire space between the plates. If the capacitor remains connected to a battery of voltage $V_0$, what is the new charge stored on the capacitor? Solution: Initial capacitance (air): $C_0$. Initial voltage: $V_0$. Since the capacitor remains connected to the battery, the voltage across it remains constant at $V_0$. When a dielectric slab of constant $k$ fills the space, the new capacitance becomes $C_{new} = kC_0$. Given $k=4$, so $C_{new} = 4C_0$. The new charge stored is $Q_{new} = C_{new} \cdot V_0 = (4C_0) \cdot V_0 = 4 (C_0 V_0)$. If $Q_0 = C_0 V_0$ was the initial charge, then $Q_{new} = 4Q_0$.