Electric Current — Class 10 Physical Science
"Electricity is the backbone of modern civilisation. It lights our homes, charges our phones, and powers our lives."
1. Basic Quantities
| Quantity | Symbol | Unit | Definition |
|---|---|---|---|
| Charge | Q | Coulomb (C) | Property of matter. 1 electron charge = −1.6×10⁻¹⁹ C. |
| Current | I | Ampere (A) | Rate of flow of charge. I = Q/t. 1 A = 1 C/s. |
| Potential Difference | V | Volt (V) | Work done per unit charge. V = W/Q. 1 V = 1 J/C. |
2. Ohm's Law: V = IR
At constant temperature, the current through a conductor is DIRECTLY PROPORTIONAL to the potential difference across it. V-I graph is a STRAIGHT LINE through the origin (for ohmic conductors — metals). 'Non-ohmic conductors (semiconductors, electrolytes) do NOT obey Ohm's Law — their V-I graph is curved.'
3. Resistance: R = V/I. Unit: Ohm (Ω).
Factors: R ∝ LENGTH (longer wire = more resistance). R ∝ 1/AREA (thicker wire = less resistance — more space for electrons). R depends on MATERIAL — resistivity (ρ). R depends on TEMPERATURE — for metals: R increases with temperature.
Resistivity: R = ρL/A. ρ depends ONLY on the material — NOT on dimensions. Unit: Ω·m. Copper: ρ ≈ 1.7×10⁻⁸ Ω·m (very low — excellent conductor). Nichrome: ρ ≈ 100×10⁻⁸ Ω·m (much higher — used in heating elements).
4. Series and Parallel Circuits
| Series | Parallel | |
|---|---|---|
| Current (I) | SAME everywhere | DIVIDES at junctions. I_total = I₁+I₂+I₃ |
| Voltage (V) | DIVIDES. V_total = V₁+V₂+V₃ | SAME across each branch |
| Equivalent Resistance | R_eq = R₁+R₂+R₃ (always LARGER than largest individual) | 1/R_eq = 1/R₁+1/R₂+1/R₃ (R_eq is SMALLER than smallest individual) |
| If one device fails | ALL go OFF (circuit broken) | Others stay ON (independent paths) |
| Domestic wiring | NOT used | USED — each appliance gets full voltage |
5. Electrical Power: P = VI = I²R = V²/R
Unit: WATT (W). Energy consumed: E = P × t. Commercial unit: 1 kWh = 1 "unit" = 3.6 × 10⁶ J. 'Your electricity meter reads in kWh (units). A 1000W heater running for 1 hour = 1 unit.'
6. Joule's Heating: H = I²Rt
When current flows through a resistor: electrical energy → HEAT. Applications: Electric iron. Heater. Toaster. Electric bulb (filament heats to ~2500°C → glows WHITE). Disadvantage: Wasted energy in transmission lines. 'Power is transmitted at HIGH VOLTAGE to REDUCE current — because H ∝ I², lower current = MUCH lower heat loss.'
7. Fuse
A THIN wire with LOW melting point (tin-lead alloy). Connected in SERIES with the LIVE wire. If current exceeds RATING → fuse MELTS → circuit BREAKS. 'The fuse is a SACRIFICIAL PROTECTOR. It DIES to save the appliance — and YOU.'
8. Common Mistakes
- 'Series: voltage is same' — WRONG. Voltage is SAME in PARALLEL. In SERIES, current is same.
- 'More resistance = more current' — I = V/R. MORE resistance = LESS current (for same voltage).
- Using V²/R for power in series: Be CAREFUL which V you use — the voltage ACROSS that component, not the total.
9. AP SSC Exam Focus
| Topic | Marks |
|---|---|
| Ohm's Law and V-I graph | 3-4 |
| Series/Parallel problems | 4-5 |
| Power and energy (kWh) | 3-4 |
| Joule's heating | 2-3 |
| Fuse — principle and use | 2-3 |
10. Worked Numerical Problems
Ohm's Law Numericals
Example 1: A potential difference of 12 V is applied across a resistor. A current of 3 A flows. Find the resistance. Solution: R = V/I = 12/3 = 4 Ω.
Example 2: An electric iron draws a current of 5 A when connected to 220 V. What is its resistance? Solution: R = V/I = 220/5 = 44 Ω.
Resistivity Numericals
Example 3: A copper wire of length 2 m and area of cross-section 1.7 × 10⁻⁶ m² has resistivity 1.7 × 10⁻⁸ Ω·m. Find its resistance. Solution: R = ρL/A = (1.7 × 10⁻⁸ × 2) / (1.7 × 10⁻⁶) = 0.02 Ω.
Example 4: A wire of resistance 10 Ω is stretched to DOUBLE its length. What is the new resistance? Solution: When length doubles, area HALVES (volume constant). R ∝ L/A. New R = 10 × (2L/L) / (A/2A) = 10 × 2 × 2 = 40 Ω. 'Stretching a wire to n times its length MULTIPLIES resistance by n².'
Series and Parallel Numericals
Example 5: Three resistors of 2 Ω, 3 Ω, and 5 Ω are connected in SERIES. Find the equivalent resistance. Solution: R_eq = R₁ + R₂ + R₃ = 2 + 3 + 5 = 10 Ω.
Example 6: Three resistors of 2 Ω, 3 Ω, and 6 Ω are connected in PARALLEL. Find the equivalent resistance. Solution: 1/R_eq = 1/2 + 1/3 + 1/6 = 3/6 + 2/6 + 1/6 = 6/6 = 1. R_eq = 1 Ω. 'Notice: R_eq in parallel is LESS than the SMALLEST individual resistor.'
Example 7: A 4 Ω and a 6 Ω resistor are connected in parallel. This combination is connected in SERIES with a 2 Ω resistor. Find the total resistance. Solution: R_parallel = (4×6)/(4+6) = 24/10 = 2.4 Ω. R_total = 2.4 + 2 = 4.4 Ω.
Power and Energy Numericals
Example 8: A 100 W bulb is used for 5 hours daily. Find energy consumed in kWh in 30 days. Solution: Daily energy = 100 × 5 = 500 Wh = 0.5 kWh. Monthly = 0.5 × 30 = 15 kWh (units). Cost at ₹8/unit = 15 × 8 = ₹120.
Example 9: A heater of resistance 50 Ω draws 4 A current. Find its power. Solution: P = I²R = (4)² × 50 = 16 × 50 = 800 W.
Joule's Heating Numericals
Example 10: A current of 5 A flows through a heater of resistance 20 Ω for 30 minutes. Find the heat produced. Solution: H = I²Rt = 5² × 20 × (30×60) = 25 × 20 × 1800 = 900,000 J = 900 kJ.
11. MCB — Miniature Circuit Breaker
MCB is a MODERN alternative to the fuse. It uses an ELECTROMAGNET or BIMETALLIC strip. When current exceeds the rating: the electromagnetic coil PULLS a switch → circuit BREAKS. 'Unlike a fuse, an MCB does NOT need replacement. Just FLIP the switch back ON — once the fault is fixed.'
Advantages over Fuse: Reusable (no replacement needed). Faster tripping. More reliable. Can be reset remotely.
12. Factors Affecting Resistance — Detailed
Length (L): R ∝ L. 'Electrons collide MORE with atoms in a LONGER wire — more resistance.'
Cross-sectional Area (A): R ∝ 1/A. 'A THICKER wire provides MORE paths for electrons — LESS resistance. That's why thick wires are used for high-current appliances.'
Material (ρ): Each material has its OWN resistivity. Silver (1.6×10⁻⁸ Ω·m) is the BEST conductor. Copper (1.7×10⁻⁸) is a close second — and much CHEAPER. That's why copper wires are used everywhere.
Temperature: For metals, R INCREASES with temperature (atoms vibrate MORE → impede electron flow). For semiconductors (silicon, germanium), R DECREASES with temperature.
13. Domestic Electric Circuits — Detailed
Ring Circuit: A LOOP of wire starting from the mains, going around the house, and RETURNING to the mains. Each socket is connected to the ring. 'The ring circuit uses THINNER wire than a radial circuit — because current is shared between two paths.'
3-Pin Plug: LIVE (brown — carries current IN). NEUTRAL (blue — returns current). EARTH (green/yellow — SAFETY — connected to metal casing). 'The earth pin is LONGER and THICKER — it makes contact FIRST when plugging in, ensuring the appliance is earthed BEFORE power reaches it.'
Overload: Too many appliances on one circuit → current exceeds safe limit → wires OVERHEAT → fire risk. 'This is WHY fuses/MCBs are essential — they TRIP when current exceeds rating and PREVENT fire.'
14. Self-Test
Q1: A wire of resistance 5 Ω is cut into 5 EQUAL pieces. What is the resistance of each piece? If all 5 pieces are connected in parallel, find the equivalent resistance. A1: Each piece = 5/5 = 1 Ω. In parallel: 1/R_eq = 1+1+1+1+1 = 5. R_eq = 1/5 = 0.2 Ω.
Q2: Why is the heating element of a toaster made of nichrome and NOT copper? A2: Nichrome has HIGH resistivity (generates MUCH heat) and HIGH melting point (doesn't melt at operating temperature). Copper has very LOW resistivity — it would produce almost NO heat.
Q3: Two bulbs of 60 W and 100 W are connected in SERIES to a 220 V supply. Which will glow BRIGHTER? A3: The 60 W bulb has HIGHER resistance (R = V²/P). In series, CURRENT is same. Power = I²R. So the bulb with HIGHER resistance (60 W) will glow BRIGHTER.
Q4: Why is MCB preferred over a fuse in modern homes? A4: MCB can be REUSED (just flip the switch). Fuse needs REPLACEMENT (wire melts). MCB trips FASTER. MCB is more accurate and reliable.
Q5: A current of 0.5 A flows through a bulb for 2 minutes. Calculate the charge that flows through it. A5: Q = I × t = 0.5 × (2 × 60) = 0.5 × 120 = 60 C.
Q6: Explain why the resistance of a wire increases with temperature. A6: At higher temperature, atoms in the wire VIBRATE MORE. These vibrations COLLIDE with flowing electrons → impede their motion → more resistance.
Q7: Find the current drawn by an electric heater of power 1500 W when connected to 220 V. A7: P = VI → I = P/V = 1500/220 = 6.82 A (approximately).
