Current Electricity + Capacitance Notes and Important Questions for NEET by PW

Current Electricity deals with the flow of charge, resistance, potential difference and circuit analysis. Capacitance focuses on charge storage, electric fields and energy. However, solving questions from these chapters can be challenging when you need to choose the right formula, analyse complex circuit networks or determine how a capacitor behaves under different conditions.

The Current Electricity + Capacitance NEET One-Shot Video by PW focuses on concept-building and question practice across different topics. Use these notes to revise key concepts, recall formulas and practise numerical problems before your next revision session.

Important Notes: Current Electricity and Capacitance

1. Kirchhoff’s Laws and Current Division

• Junction law: Total current entering a junction equals total current leaving it.

• Loop law: The algebraic sum of potential differences around a closed loop is zero.

• Current division: For two parallel resistors, current through one branch is inversely proportional to its resistance.

For two parallel resistors R1 and R2 carrying total current I:

I1 = [R2 / (R1 + R2)] × I

I2 = [R1 / (R1 + R2)] × I

The branch with lower resistance carries more current.

2. Node-Potential Method and Wheatstone Bridge

• In the node-potential method, assign a reference point 0 V and determine the potentials of other nodes.

• Current through a resistor is the potential difference divided by its resistance.

• A Wheatstone bridge is balanced when R1 × R4 = R2 × R3.

• In a balanced bridge, the central branch has zero potential difference, so no current flows through it.

• Points connected by ideal wires have the same potential. Use this principle to simplify symmetric resistor networks.

An ideal ammeter has zero resistance and behaves like an ideal wire.

3. Cells and Maximum Power Transfer

For a cell with EMF E, internal resistance r and current I:

• Discharging: Terminal voltage = E - Ir

• Charging: Terminal voltage = E + Ir

For N identical cells:

Maximum power is transferred when external resistance equals internal resistance.

Maximum power = E² / 4r

Efficiency = [External resistance / (External resistance + Internal resistance)] × 100

At maximum power transfer, efficiency is 50%.

4. Bulbs, Meter Bridge and Galvanometer

For a bulb rated at voltage V and power P:

Resistance = V² / P

• Bulbs in series: The bulb with higher resistance dissipates more power.

• Bulbs in parallel: The bulb with lower resistance dissipates more power.

• Meter bridge: It works on the Wheatstone bridge principle. For a uniform wire at the balance point, R1 / R2 = l1 / l2.

• Galvanometer to ammeter: Connect a low-resistance shunt in parallel.

Shunt resistance Rs = (Ig × G) / (I - Ig)

• Galvanometer to voltmeter: Connect a high resistance in series.

Required series resistance = V / Ig - G

Here, G is galvanometer resistance, Ig is full-scale galvanometer current and I is the desired ammeter range.

5. Drift Velocity, Current Density and Resistance

Important formulas:

• Current: I = dq / dt

• Current due to charge carriers: I = neAvd

• Current density: J = I / A

• Current density relation: J = σE

• Resistance: R = ρL / A

In a steady conductor, current remains the same through different cross-sections. Drift velocity is greater in the narrower section when the current and carrier density remain constant.

For metallic conductors, resistance generally increases with temperature. For semiconductors, resistance generally decreases with temperature.

6. Capacitance and Dielectric Effects

Capacitance is the ratio of charge to potential difference.

C = Q / V

For a parallel-plate capacitor without a dielectric:

C = ε₀A / d

When a dielectric of constant K completely fills the gap:

C' = KC

The effect of inserting a dielectric depends on whether the battery remains connected.

Remember: With the battery disconnected, charge remains constant. With the battery connected, potential difference remains constant.

7. Capacitor Combinations and Energy

For capacitors in series:

1 / Ceq = 1 / C1 + 1 / C2 + 1 / C3

For capacitors in parallel:

Ceq = C1 + C2 + C3

Energy stored in a capacitor:

U = 1/2 CV² = Q² / (2C) = 1/2 QV

Energy density of an electric field:

u = 1/2 ε₀E²

Charge on a capacitor:

Q = C × ΔV

Dielectric slabs arranged along the plate separation act like capacitors in series. Dielectric regions dividing the plate area act like capacitors in parallel.

8. Charge Sharing and Capacitor Networks

When two capacitors are connected with like polarities and reach a common potential:

Final potential = (C1V1 + C2V2) / (C1 + C2)

Energy loss during charge redistribution:

Energy loss = 1/2 × [C1C2 / (C1 + C2)] × (V1 - V2)²

To simplify capacitor networks, identify points joined by ideal wires, locate equal-potential points and remove capacitors connected between them. Then apply series and parallel combinations.

Important Questions: Current Electricity and Capacitance

1. A junction has currents of 3 A and 5 A entering it. Find the current leaving the junction.

Answer: By Kirchhoff’s junction law, current leaving = 3 + 5 = 8 A.

2. Two resistors of 2 ohms and 4 ohms are connected in parallel. If the total current is 6 A, find the current through the 2-ohm resistor.

Answer: Current through the 2-ohm resistor = [4 / (2 + 4)] × 6 = 4 A.

3. A Wheatstone bridge has resistances R1 = 2 ohms, R2 = 4 ohms and R3 = 3 ohms. Find R4 for a balanced bridge.

Answer: For balance, R1 × R4 = R2 × R3.

2 × R4 = 4 × 3

R4 = 6 ohms.

4. Three identical cells, each with EMF 2 V and internal resistance 0.5 ohm, are connected in series. Find their equivalent EMF and internal resistance.

Answer: Equivalent EMF = 3 × 2 = 6 V.

Equivalent internal resistance = 3 × 0.5 = 1.5 ohms.

5. A cell has EMF 12 V and internal resistance 2 ohms. Find the maximum power delivered to the external resistor.

Answer: Maximum power occurs when external resistance = internal resistance = 2 ohms.

Maximum power = E² / (4r) = 12² / (4 × 2) = 18 W.

6. A bulb is rated 100 W, 200 V. Find its resistance at the rated operating conditions.

Answer: Resistance = V² / P = 200² / 100 = 400 ohms.

7. A meter bridge has balance lengths of 40 cm and 60 cm. If the wire is uniform, find the ratio R1 / R2.

Answer: R1 / R2 = 40 / 60 = 2 / 3.

8. A galvanometer has resistance 10 ohms and full-scale current 1 mA. Calculate the shunt resistance needed to convert it into a 1 A ammeter.

Answer: Shunt resistance = (Ig × G) / (I - Ig)

= (0.001 × 10) / (1 - 0.001)

Approximately 0.010 ohm.

9. A conductor carries a current of 4 A through a cross-sectional area of 2 × 10⁻⁶ m². Calculate its current density.

Answer: J = I / A = 4 / (2 × 10⁻⁶) = 2 × 10⁶ A/m².

10. A conductor carries a steady current through two sections. The second section has half the cross-sectional area of the first. Compare their drift velocities, assuming the carrier density is unchanged.

Answer: Since I = neAvd and the current remains constant, drift velocity is inversely proportional to area.

The second section has half the area, so its drift velocity is twice that of the first section.

11. A parallel-plate capacitor has capacitance 4 μF. A dielectric of constant 3 completely fills the gap. Find the new capacitance.

Answer: New capacitance = KC = 3 × 4 = 12 μF.

12. A charged capacitor is disconnected from its battery before a dielectric of constant 2 is inserted completely. If its initial charge is 20 μC and potential is 10 V, find its new potential and stored energy relative to the initial energy.

Answer: The charge remains constant because the battery is disconnected.

New potential = 10 / 2 = 5 V.

New energy = Initial energy / 2.

13. Capacitors of 3 μF and 6 μF are connected in series. Find their equivalent capacitance.

Answer: Ceq = (3 × 6) / (3 + 6) = 2 μF.

14. A 2 μF capacitor charged to 6 V is connected with like polarity to an uncharged 4 μF capacitor. Find the common final potential and energy loss.

Answer: Final potential = (2 × 6 + 4 × 0) / (2 + 4) = 2 V.

Initial energy = 1/2 × 2 μF × 6² = 36 μJ.

Final energy = 1/2 × 6 μF × 2² = 12 μJ.

Energy loss = 36 - 12 = 24 μJ.

15. A capacitor of 5 μF is charged to 10 V. Calculate its stored energy.

Answer: U = 1/2 CV² = 1/2 × 5 × 10⁻⁶ × 10² = 2.5 × 10⁻⁴ J.

16. A cell of EMF 10 V and internal resistance 2 ohms is connected to an external resistor of 3 ohms. Find the current and terminal voltage of the cell.

Answer: Total resistance = 3 + 2 = 5 ohms.

Current = E / (R + r) = 10 / 5 = 2 A.

Terminal voltage = E - Ir = 10 - (2 × 2) = 6 V.

17. A galvanometer has resistance 20 ohms and full-scale current 2 mA. What resistance must be connected in series to convert it into a voltmeter with a 10 V range?

Answer: Required series resistance = V / Ig - G.

= 10 / 0.002 - 20

= 5,000 - 20 = 4,980 ohms.

18. Two capacitors of 4 μF and 12 μF are connected in parallel across a 6 V battery. Find the equivalent capacitance and total charge supplied by the battery.

Answer: Equivalent capacitance = 4 + 12 = 16 μF.

Total charge = Ceq × V = 16 × 6 = 96 μC.

19. A parallel-plate capacitor has plate area 0.02 m² and separation 1 mm. Calculate its capacitance in air, taking ε₀ = 8.85 × 10⁻¹² F/m.

Answer: C = ε₀A / d.

= (8.85 × 10⁻¹² × 0.02) / (1 × 10⁻³)

= 1.77 × 10⁻¹⁰ F.

20. Two capacitors of 3 μF and 6 μF are connected in parallel across a 12 V battery. Calculate the charge on each capacitor and the total stored energy.

Answer: In parallel, both capacitors have the same potential difference of 12 V.

Charge on 3 μF capacitor = 3 × 12 = 36 μC.

Charge on the 6 μF capacitor = 6 × 12 = 72 μC.

Total capacitance = 3 + 6 = 9 μF.

Total energy = 1/2 × 9 × 10⁻⁶ × 12² = 6.48 × 10⁻⁴ J.

Watch the Current Electricity + Capacitance FULL CHAPTER || NEET one-shot video by PW to practise more varied questions and strengthen your confidence in applying formulas, analysing circuit networks and solving capacitor problems. Attempt the questions independently, review the solutions and revisit the concepts where you need more practice.

Current Electricity and Capacitance require a clear understanding of circuit rules, resistance, capacitor combinations and energy relations. Revise the key formulas, practise numerical problems and check your calculations carefully. The PW one-shot video can help you build concepts and practise questions from different topics.

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