Electricity — NCERT Solutions, AI Tutor & Practice

Definition: Electric current is the rate of flow of electric charge through a conductor.

$I = \frac{Q}{t}$

SI unit: ampere (A), where $1\,\text{A} = 1\,\text{C/s}$.

Calculation:

• Charge $Q = 60\,\text{C}$
• Time $t = 2\,\text{min} = 2 \times 60 = 120\,\text{s}$

$I = \frac{60}{120} = 0.5\,\text{A}$

Ohm's law: At constant temperature, the current flowing through a conductor is directly proportional to the potential difference across its ends.

$V \propto I \implies V = IR$

where $R$ is the resistance (constant for an ohmic conductor).

V–I graph: A straight line passing through the origin.

 V
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 |      /
 |    /
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 |/________ I

Slope: The slope of the V–I graph equals the resistance $R$ of the conductor, since $R = V/I$.

Formula: $R = \rho \dfrac{l}{A}$

When the wire is stretched, its volume stays constant: $V = A \times l = A' \times l'$.

If length doubles, $l' = 2l$, so: $A' = \frac{A \times l}{l'} = \frac{A \times l}{2l} = \frac{A}{2}$

New resistance: $R' = \rho \frac{l'}{A'} = \rho \frac{2l}{A/2} = 4 \times \rho \frac{l}{A} = 4R$

$R' = 4 \times 20 = 80\,\Omega$

The resistance of a conductor depends on:

1. Length ($l$): $R \propto l$ — longer wire has more resistance.
2. Cross-sectional area ($A$): $R \propto \dfrac{1}{A}$ — thicker wire has less resistance.
3. Nature of material (resistivity $\rho$)
4. Temperature (resistance generally increases with temperature for metals).

Relation: $R = \rho \frac{l}{A}$

Resistivity ($\rho$): The resistance of a conductor of unit length and unit cross-sectional area. Its SI unit is ohm-metre ($\Omega\,\text{m}$). It is a property of the material and does not depend on the dimensions of the conductor.

(a) Total resistance in series: $R_s = R_1 + R_2 + R_3 = 5 + 10 + 15 = 30\,\Omega$

(b) Current in the circuit (same through all resistors in series): $I = \frac{V}{R_s} = \frac{6}{30} = 0.2\,\text{A}$

Equivalent resistance in parallel: $\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{6} + \frac{1}{3} = \frac{1}{6} + \frac{2}{6} = \frac{3}{6} = \frac{1}{2}$

$R_p = 2\,\Omega$

Total current from supply: $I = \frac{V}{R_p} = \frac{12}{2} = 6\,\text{A}$

Joule's law of heating: The heat produced in a resistor is directly proportional to (i) the square of the current $I^2$, (ii) the resistance $R$, and (iii) the time $t$ for which the current flows.

$H = I^2 R t$

Since $V = IR$, we can also write $H = VIt$.

Calculation:

• $V = 220\,\text{V}$, $I = 5\,\text{A}$, $t = 30\,\text{s}$

$H = VIt = 220 \times 5 \times 30 = 33000\,\text{J}$

$H = 3.3 \times 10^4\,\text{J}$

(a) Current: $P = VI$ $I = \frac{P}{V} = \frac{60}{220} \approx 0.27\,\text{A}$

(b) Resistance: $P = \dfrac{V^2}{R} \implies R = \dfrac{V^2}{P}$ $R = \frac{(220)^2}{60} = \frac{48400}{60} \approx 806.7\,\Omega$

(c) Energy consumed:

• Power $= 60\,\text{W} = 0.06\,\text{kW}$
• Total time $= 5 \times 30 = 150\,\text{hours}$

$E = P \times t = 0.06 \times 150 = 9\,\text{kWh}$

Tungsten for bulb filaments:

• It has a very high melting point ($\approx 3380^\circ\text{C}$), so it does not melt at the high operating temperature.
• It has high resistivity, so it gets sufficiently hot to glow and emit light.
• It can be drawn into thin wires and retains strength at high temperature.

Alloys for heating elements (e.g. nichrome):

• Alloys have higher resistivity than pure metals, producing more heat.
• They have a high melting point, allowing operation at high temperatures.
• They do not oxidise (burn) easily at high temperatures, so they last longer.

Step 1: Parallel combination of 4 Ω and 12 Ω: $\frac{1}{R_p} = \frac{1}{4} + \frac{1}{12} = \frac{3}{12} + \frac{1}{12} = \frac{4}{12} = \frac{1}{3}$ $R_p = 3\,\Omega$

Step 2: Series with 5 Ω: $R_{total} = R_p + 5 = 3 + 5 = 8\,\Omega$

Step 3: Main current: $I = \frac{V}{R_{total}} = \frac{12}{8} = 1.5\,\text{A}$

Household appliances are connected in parallel because:

1. Same voltage: Each appliance receives the full supply voltage (220 V) required for proper working.

2. Independent operation: If one appliance is switched off or fails, the others continue to work, since each has its own separate path.

3. Different currents: Each appliance draws current according to its own resistance/power rating, independent of the others.

4. Lower total resistance: The overall resistance of the circuit decreases, allowing enough current for all appliances.

In a series connection, the same current would flow through all, the voltage would be divided, and failure of one device would break the whole circuit.

(a) Resistance: By Ohm's law, $R = \frac{V}{I} = \frac{6}{0.5} = 12\,\Omega$

(b) Power consumed: $P = VI = 6 \times 0.5 = 3\,\text{W}$

(Check: $P = I^2R = (0.5)^2 \times 12 = 0.25 \times 12 = 3\,\text{W}$ ✓)