Current Electricity

Class 10 Physics • Chapter 08 (Deep Detail)

1. Ohm's Law

Statement: The current ($I$) flowing in a conductor is directly proportional to the potential difference ($V$) applied across its ends, provided physical conditions (temperature, etc.) remain constant.

FORMULA

$$ V = I \times R $$

Where $R$ is the Resistance of the conductor.

Ohmic Conductors: Obey Ohm's Law (e.g., all metals). V-I graph is a straight line passing through origin.

Non-Ohmic Conductors: Do not obey Ohm's Law (e.g., diode, filament bulb). V-I graph is a curve.

Practice Q1: Ohm's Law Check

Q: Is a filament lamp an ohmic resistor?

Ans: No. Its resistance increases with temperature (Non-Ohmic).

Practice Q2: V-I Graph

CONCEPTUAL Draw the V-I graph for an Ohmic conductor and a Non-Ohmic conductor (Diode).

Ans: Ohmic: Straight line passing through origin. Non-Ohmic: Curve (not straight).

2. Resistance

The obstruction offered to the flow of current.

Factors affecting Resistance:

Length ($l$): $R \propto l$.
Area of Cross-section ($A$): $R \propto \frac{1}{A}$.
Nature of Material: Depends on resistivity ($\rho$).
Temperature: Resistance of metals increases with temperature.

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

$\rho$ (Rho) is Specific Resistance or Resistivity.

Unit of R: Ohm ($\Omega$).

Unit of $\rho$: Ohm-metre ($\Omega \cdot m$).

Practice Q3: Resistivity

NUMERICAL A wire of length 2 m and area $10^{-6} \text{ m}^2$ has a resistance of $0.04 \Omega$. Find its resistivity.

Solution:

$\rho = \frac{R A}{l} = \frac{0.04 \times 10^{-6}}{2}$.

$\rho = 0.02 \times 10^{-6} = 2 \times 10^{-8} \Omega \text{ m}$.

Practice Q4: Resistance Dependence

Q: If a wire is doubled on itself, what happens to its resistance?

Ans: Length halves ($l/2$), Area doubles ($2A$). $R' \propto \frac{l/2}{2A} = \frac{1}{4}R$. Resistance becomes one-fourth.

Practice Q5: Stretching Wire

THINKING A wire of resistance $10 \Omega$ is stretched to double its length. What is its new resistance?

Solution:

When length doubles ($l' = 2l$), area becomes half ($A' = A/2$).

$R' = n^2 R = 2^2 \times 10 = 40 \Omega$.

3. EMF and Internal Resistance

EMF (Electromotive Force) $\varepsilon$: Potential difference when circuit is OPEN (No current drawn).

Terminal Voltage ($V$): Potential difference when circuit is CLOSED (Current drawn).

Voltage Drop: $V = \varepsilon - I r$ (where $r$ is internal resistance of cell).

Practice Q6: Voltage Drop Graph

[IMAGE PLACEHOLDER: V VS I GRAPH]

Graph of Terminal Voltage (V) on Y-axis vs Current (I) on X-axis. Straight line with negative slope. Y-intercept is EMF. Slope is -r. (Invert in dark mode).

GRAPH ANALYSIS From the graph of V vs I for a cell, how can you determine EMF and internal resistance?

Ans: 1. Y-intercept = EMF ($\varepsilon$). 2. Slope = Internal Resistance (-r).

Practice Q7: Internal Resistance

NUMERICAL A cell of EMF 1.5 V and internal resistance $1 \Omega$ is connected to two resistors of $4 \Omega$ and $20 \Omega$ in series. Calculate voltage drop across the $20 \Omega$ resistor.

Solution:

Total External Resistance $R = 4 + 20 = 24 \Omega$. Total $R_{net} = 24 + 1 = 25 \Omega$.

Current $I = 1.5/25 = 0.06 \text{ A}$. $V_{20} = I \times 20 = 0.06 \times 20 = 1.2 \text{ V}$.

4. Combination of Resistors

[IMAGE PLACEHOLDER: SERIES VS PARALLEL]

Use circuit diagrams to show Resistors in Series vs Parallel. Label current 'I' path and Voltage 'V' drops clearly.

Series Combination

Current is same in all resistors.
Voltage divides ($V = V_1 + V_2$).
Total $R_s = R_1 + R_2 + R_3$.
Result: Resistance increases to maximum.

Parallel Combination

Voltage is same across all resistors.
Current divides ($I = I_1 + I_2$).
$\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \dots$
Result: Resistance decreases to minimum.

Practice Q8: Mixed Circuit

[IMAGE PLACEHOLDER: MIXED RESISTORS CIRCUIT]

Circuit Diagram: 6V Battery. Two resistors R1(3 ohm) and R2(6 ohm) in Parallel. This group is in series with R3(2 ohm). Ammeter in series. Voltmeter across Parallel group. (Invert in dark mode).

NUMERICAL Calculate: (a) Total Resistance, (b) Total Current, (c) Reading of Voltmeter connected across the parallel combination ($3 \Omega$ and $6 \Omega$).

Solution:

(a) $1/R_p = 1/3 + 1/6 = 3/6 = 1/2$. So $R_p = 2 \Omega$. Total $R = 2 + 2 = 4 \Omega$.

(b) Current $I = V/R = 6/4 = 1.5 \text{ A}$.

Practice Q9: Combination of Resistors

NUMERICAL Two resistors of $4 \Omega$ and $6 \Omega$ are connected in parallel. This combination is connected in series with a $2.6 \Omega$ resistor and a battery of 6V of negligible internal resistance. Calculate current.

Solution:

1. Parallel: $R_p = (4 \times 6)/(4+6) = 24/10 = 2.4 \Omega$.

2. Total $R = 2.4 + 2.6 = 5 \Omega$.

3. current $I = V/R = 6/5 = 1.2 \text{ A}$.

5. Electrical Energy and Power

POWER FORMULA

Basic: $P = V \times I$

Using Ohm's Law ($V=IR$): $P = I^2 R$ (Heat loss in series).

Using Ohm's Law ($I=V/R$): $P = \frac{V^2}{R}$ (Appliances in parallel).

HEAT ENERGY

$$ H = I^2 R t $$

(Joule's Law of Heating)

$$ H = V I t $$ $$ H = V I t $$

Commercial Unit of Energy: kWh (Board of Trade Unit).

Practice Q10: Bulb Rating

NUMERICAL An electric bulb is rated '220 V, 100 W'. (a) What is its resistance? (b) What is the safe limit of current?

Solution:

(a) $R = V^2 / P = (220 \times 220) / 100 = 484 \Omega$.

(b) $I = P / V = 100 / 220 \approx 0.45 \text{ A}$.

Practice Q11: Bill Calculation

NUMERICAL An electric heater of 2 kW is used for 4 hours daily. Calculate the cost for 30 days if rate is ₹5 per unit.

Solution: Energy/day = $2 \times 4 = 8 \text{ kWh}$. Total Energy = $240 \text{ units}$. Cost = $240 \times 5 = 1200 \text{ ₹}$.

Practice Q12: Electric Power

CONCEPTUAL Two bulbs 60W and 100W are connected in series. Which one glows brighter?

Solution: In Series, $I$ is same. $P = I^2 R$. $R_{60W} > R_{100W}$. So 60W bulb glows brighter.