Class 11 Physics DC Circuit Notes

Unit 4

Electricity and Magnetism

Class 11 Physics

Chapter 23

DC Circuits

Class 11 Physics – DC Circuits Notes PDF

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Chapter Overview

Direct-current circuits describe the steady flow of electric charge through conductors and circuit components. The chapter connects the microscopic motion of charge carriers with measurable quantities such as current, resistance, voltage, electromotive force and electrical power.

The official scope includes electric current and drift velocity, Ohm’s law, resistance/resistivity/conductivity, current–voltage characteristics, series and parallel resistors, potential dividers, electromotive force with internal resistance, and electrical work and power.

23.1 Electric Current and Drift Velocity

Electric Current

Electric current is the rate of flow of electric charge through a cross-section of a conductor.

I = dQ/dt    or for steady current    I = Q/t

The SI unit is the ampere (A), where 1 A = 1 C s−1.

Conventional Current and Electron Flow

Conventional current is defined in the direction positive charge would move. In metallic conductors, electrons drift in the opposite direction.

Diagram 1 — Conventional Current and Electron Drift

Conventional current I Electron drift

In a metal, electron drift is opposite to the conventional-current direction.

Drift Velocity

Without an electric field, conduction electrons move randomly and have no net average motion. When an electric field is applied, they acquire a small average drift velocity vd.

In time t, charge carriers within a cylinder of length vdt cross area A. If n is the number of free charge carriers per unit volume and q is the magnitude of charge per carrier:

Number crossing = nAvdt
Q = nqAvdt
I = nqAvd

For electrons, q = e in magnitude:

I = neAvd

Diagram 2 — Drift-Velocity Derivation

cross-section Alength = vdt carrier driftQ = nqAvdt

Current equals the charge density per unit volume times cross-sectional area and drift speed.

23.2 Ohm’s Law, Resistance, Resistivity and Conductivity

Ohm’s Law

For an ohmic conductor maintained at constant physical conditions, current is directly proportional to potential difference:

V ∝ I   ⇒   V = IR

Resistance is:

R = V/I

Resistance of a Uniform Conductor

R = ρL/A

where ρ is resistivity, L is length and A is cross-sectional area.

Resistivity is a material property with SI unit Ω m. Conductivity σ is the reciprocal:

σ = 1/ρ
QuantitySymbolDepends mainly onSI unit
ResistanceRMaterial, length, area, temperatureΩ
ResistivityρMaterial and temperatureΩ m
ConductivityσMaterial and temperatureS m⁻¹

Diagram 3 — Factors Affecting Resistance

Long thin conductor Short thick conductor R ∝ LR ∝ 1/AR = ρL/A

For the same material and temperature, increasing length increases resistance while increasing area lowers it.

23.3 Current–Voltage Characteristics: Ohmic and Non-Ohmic Devices

An ohmic conductor has a linear V–I relation over the range where its physical conditions stay constant. A non-ohmic device does not maintain a constant V/I ratio.

Diagram 4 — Ohmic and Non-Ohmic I–V Characteristics

Ohmic resistorVI Non-ohmic exampleVI

A straight line through the origin indicates constant resistance; a curved characteristic indicates resistance varies with operating conditions.

Examples: A metallic resistor operated over a modest temperature range can be approximately ohmic. A filament lamp and many semiconductor devices are non-ohmic.

23.4 Resistors in Series and Parallel

Series Combination

The same current flows through all series resistors, while potential differences add:

V = IR₁ + IR₂ + … = I(R₁+R₂+…)
Req = R₁ + R₂ + R₃ + …

Diagram 5 — Resistors in Series

R₁R₂R₃Req = R₁ + R₂ + R₃

Series resistance is greater than any individual resistance.

Parallel Combination

Each branch has the same potential difference, while currents add:

I = V/R₁ + V/R₂ + …
1/Req = 1/R₁ + 1/R₂ + 1/R₃ + …

For two resistors:

Req = R₁R₂/(R₁+R₂)

Diagram 6 — Resistors in Parallel

R₁R₂R₃1/Req = 1/R₁ + 1/R₂ + 1/R₃

Parallel equivalent resistance is less than the smallest branch resistance.

23.5 Potential Divider

A potential divider uses resistors in series to obtain a chosen fraction of an applied voltage.

For R₁ and R₂ in series across Vin, with Vout taken across R₂ and with negligible loading:

Vout = Vin R₂/(R₁+R₂)

Diagram 7 — Potential Divider

R₁R₂VoutVin across R₁ + R₂

With no significant load attached, the output voltage is the fraction of the total voltage appearing across R₂.

Loading effect: If a low-resistance load is connected across R₂, it changes the effective lower resistance and therefore changes Vout.

23.6 Electromotive Force and Internal Resistance

Electromotive Force

The electromotive force (emf) ε of a source is the energy supplied by the source per unit charge around the complete circuit:

ε = W/Q

Despite its name, emf is measured in volts and is not a force.

Internal Resistance

A real source has internal resistance r. If current I is delivered to an external load R:

ε = IR + Ir
ε = I(R+r)

The terminal potential difference while the source delivers current is:

V = ε − Ir = IR

Diagram 8 — Real Cell with Internal Resistance

ε r external R Vterminal = ε − Ir

Part of the source emf is lost across internal resistance when current flows.

23.7 Work and Power in Electrical Circuits

When charge Q moves through potential difference V, electrical work is:

W = VQ

Since Q = It:

W = VIt

Electrical power is:

P = W/t = VI

For a resistor using V = IR:

P = I²R = V²/R

Electrical energy dissipated in time t is:

E = Pt = VIt = I²Rt = (V²/R)t
QuantityFormulaSI unit
Electrical work / energyW = VItJ
PowerP = VIW
Resistive powerP = I²R = V²/RW

Solved Numerical Examples

Example 1 — Drift Velocity

Question: A wire carries 2.0 A. If n = 8.5×10²⁸ m⁻³, A = 1.0×10⁻⁶ m² and e = 1.60×10⁻¹⁹ C, find vd.

vd = I/(neA) ≈ 2/[8.5×10²⁸ ×1.60×10⁻¹⁹ ×10⁻⁶] ≈ 1.47×10⁻⁴ m s⁻¹

Answer: about 1.5×10⁻⁴ m s⁻¹.

Example 2 — Resistance

Question: A wire has ρ = 1.7×10⁻⁸ Ωm, L = 2.0 m and A = 1.0×10⁻⁶ m². Find R.

R = ρL/A = 3.4×10⁻² Ω

Answer: 0.034 Ω.

Example 3 — Series and Parallel

Question: Find the equivalent resistance of 6 Ω and 3 Ω in parallel.

Req = (6×3)/(6+3) = 2 Ω

Answer: 2 Ω.

Example 4 — Potential Divider

Question: R₁ = 2 kΩ and R₂ = 3 kΩ are connected across 10 V. Find the unloaded Vout across R₂.

Vout = 10×3/(2+3) = 6 V

Answer: 6 V.

Example 5 — Internal Resistance

Question: A cell of emf 1.5 V and internal resistance 0.20 Ω supplies 1.0 A. Find terminal voltage.

V = ε − Ir = 1.5 − (1.0)(0.20) = 1.30 V

Answer: 1.30 V.

Example 6 — Electrical Power

Question: A 12 V lamp draws 2 A. Find its power.

P = VI = 12×2 = 24 W

Answer: 24 W.

Important Exam Questions

Short-Answer Questions

  1. Define electric current and write its SI unit.
  2. Distinguish conventional current from electron drift.
  3. Define drift velocity.
  4. State Ohm’s law.
  5. Differentiate resistance and resistivity.
  6. Define conductivity.
  7. Distinguish ohmic and non-ohmic resistance.
  8. State the equivalent-resistance relations for series and parallel circuits.
  9. What is a potential divider?
  10. Define emf in terms of energy per unit charge.
  11. What is internal resistance?
  12. Write the three common power relations for a resistor.

Long-Answer / Derivation Questions

  1. Derive I = neAvd.
  2. Explain Ohm’s law and derive R = ρL/A from proportional relationships.
  3. Derive equivalent resistance for resistors in series and parallel.
  4. Derive the potential-divider formula.
  5. Derive V = ε − Ir for a source with internal resistance.
  6. Derive P = VI, P = I²R and P = V²/R.

Numerical Questions

  1. Find drift velocity from I, n and A.
  2. Find resistance from resistivity, length and area.
  3. Find equivalent resistance of mixed simple series/parallel networks.
  4. Calculate an unloaded potential-divider output.
  5. Find current and terminal voltage for a cell with internal resistance.
  6. Calculate electrical work and power from V, I and t.

Diagram Practice

  1. Conventional current versus electron drift.
  2. Drift-velocity cylinder.
  3. Ohmic and non-ohmic I–V graphs.
  4. Series and parallel resistor networks.
  5. Potential divider.
  6. Real cell with emf and internal resistance.

One-Minute Revision

  • I = Q/t for steady current.
  • For charge carriers, I = nqAvd.
  • Ohm’s law: V = IR under constant physical conditions.
  • R = ρL/A.
  • σ = 1/ρ.
  • Ohmic devices have linear I–V behavior over the relevant range.
  • Series: Req = ΣR.
  • Parallel: 1/Req = Σ(1/R).
  • Potential divider: Vout = VinR₂/(R₁+R₂) when unloaded.
  • Emf is source energy supplied per unit charge.
  • Terminal voltage while delivering current: V = ε − Ir.
  • Electrical work: W = VIt.
  • Electrical power: P = VI.
  • For a resistor: P = I²R = V²/R.

Syllabus Coverage Checklist

NEB/CDC Chapter 23 scopeCovered
23.1 Electric currents; drift velocity and relation with currentYes
23.2 Ohm’s law; resistance; resistivity; conductivityYes
23.3 Current–voltage relations; ohmic and non-ohmic resistanceYes
23.4 Resistances in series and parallelYes
23.5 Potential dividerYes
23.6 Electromotive force and internal resistanceYes
23.7 Work and power in electrical circuitsYes

Source handling: The original Nepal eNotes PDF remains embedded above. The typed section follows the verified NEB/CDC syllabus and is designed as a searchable, responsive study companion. Where the PDF viewer does not expose handwritten page text, the typed section is a syllabus-aligned reconstruction and is not claimed to be a word-for-word transcription.

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