Class 10 Science Notes Chapter 11 — Electricity#
Electric current#
$$I = \frac{Q}{t}$$- Electric current — the rate of flow of charge. In metal wires, the charges that move are electrons.
- SI unit: ampere (A). 1 A = 1 C/s. Smaller units: 1 mA = 10⁻³ A, 1 μA = 10⁻⁶ A.
- SI unit of charge: coulomb (C). Charge of an electron = $1.6 \times 10^{-19}$ C, so 1 C is the charge of about $6 \times 10^{18}$ electrons.
- Conventional current flows from positive to negative, opposite to the flow of electrons.
- Ammeter measures current. It is connected in series and has low resistance.
Potential difference#
$$V = \frac{W}{Q}$$- Potential difference — work done to move a unit charge from one point to another.
- SI unit: volt (V). 1 V = 1 J/C. A cell (or battery) keeps up the potential difference by chemical action.
- Voltmeter measures potential difference. It is connected in parallel and has high resistance.
Circuit symbols#
| Component | Symbol |
|---|---|
| Electric cell | ─|‖─ (long line +) |
| Battery | several cells in series |
| Plug key / switch (open, closed) | ─( )─ , ─(•)─ |
| Wire joint / crossing without joint | dot at the joint / wires crossing |
| Bulb | ⊗ in a circle |
| Resistor | ─///─ |
| Variable resistor (rheostat) | resistor with an arrow |
| Ammeter / voltmeter | Ⓐ / Ⓥ (circles) |
Ohm’s law#
At constant temperature, the current through a metal wire is directly proportional to the potential difference across it:
$$V = IR$$
- The V–I graph for a metal conductor is a straight line through the origin.
- Resistance $R$ — opposition to the flow of current. SI unit: ohm (Ω). 1 Ω = 1 V/A.
- A rheostat changes the resistance in a circuit to control the current.
Resistance and resistivity#
$$R = \rho \frac{l}{A}$$- Resistance is directly proportional to length $l$ and inversely proportional to area of cross-section $A$. It also depends on the material and temperature.
- Resistivity $\rho$ — a property of the material. It does not change with length or area, but changes with temperature. SI unit: Ω m.
| Material | Resistivity (Ω m) |
|---|---|
| Metals and alloys | $10^{-8}$ to $10^{-6}$ (low; good conductors) |
| Insulators (rubber, glass) | $10^{12}$ to $10^{17}$ (very high) |
- Silver and copper have the lowest resistivity, so copper and aluminium are used for transmission wires.
- Alloys (like nichrome) have higher resistivity than pure metals and do not oxidise (burn) easily at high temperatures. So they are used in heating elements of toasters, irons and heaters.
- Tungsten is used for bulb filaments: very high melting point (3380 °C). Bulbs are filled with inert nitrogen and argon to make the filament last longer.
Resistors in series and parallel#
| Series | Parallel | |
|---|---|---|
| Equivalent resistance | $R_s = R_1 + R_2 + R_3$ | $\dfrac{1}{R_p} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3}$ |
| Current | Same through each resistor | Divides between resistors: $I = I_1 + I_2 + I_3$ |
| Potential difference | Divides: $V = V_1 + V_2 + V_3$ | Same across each resistor |
| Equivalent value | More than the largest resistor | Less than the smallest resistor |
Why houses use parallel circuits:
- Each appliance gets the full voltage.
- Each can be switched on or off independently; one failing does not stop the others.
- Total resistance is lower, so each device draws the current it needs.
Heating effect of current#
Joule’s law of heating:
$$H = I^2 R t$$
Heat produced is proportional to the square of the current, the resistance and the time.
Applications: electric iron, toaster, oven, kettle, heater; the bulb filament; the electric fuse.
- A fuse is a wire with a low melting point (an alloy of aluminium, copper, iron, lead and so on), connected in series. If too much current flows, it melts and breaks the circuit, protecting the appliances. Fuses are rated 1 A, 2 A, 3 A, 5 A, 10 A and so on.
Electric power and energy#
$$P = VI$$$$P = I^2R$$
$$P = \frac{V^2}{R}$$
- Power — rate at which electrical energy is used. SI unit: watt (W). 1 W = 1 V × 1 A. 1 kW = 1000 W.
- Energy = power × time.
- Commercial unit of energy: the kilowatt hour (kWh), also called a “unit”. $$1 \text{ kWh} = 3.6 \times 10^6 \text{ J}$$
Common mistakes#
- Connecting the ammeter in parallel or the voltmeter in series. Ammeter in series, voltmeter in parallel.
- Saying resistivity changes with length. Resistance does; resistivity depends only on the material and temperature.
- In parallel, adding resistances directly. Add the reciprocals, then take the reciprocal of the sum.
- Using minutes or hours in $H = I^2Rt$ with joules. Time must be in seconds.
- Forgetting that kWh is a unit of energy, not power.