Electricity

The rate at which energy is delivered by a current is determined by the electric power.

Electric power is given by:
$$P = V \times I$$
where $V$ is the potential difference (voltage) and $I$ is the electric current.

It can also be expressed as:
$$P = I^2 R = \frac{V^2}{R}$$

Thus, the power (i.e., the rate of energy delivery) depends on both the potential difference across the circuit and the current flowing through it.

Given:
- Current, $I = 5\,\text{A}$
- Voltage, $V = 220\,\text{V}$
- Time, $t = 2\,\text{h} = 2 \times 3600 = 7200\,\text{s}$

Formula used:
$$P = V \times I$$
$$W = P \times t$$

Step 1: Calculate Power
$$P = 220\,\text{V} \times 5\,\text{A} = 1100\,\text{W} = 1.1\,\text{kW}$$

Step 2: Calculate Energy consumed
$$W = P \times t = 1100\,\text{W} \times 7200\,\text{s} = 7,920,000\,\text{J} = 7.92 \times 10^6\,\text{J}$$

Alternatively in kWh:
$$W = 1.1\,\text{kW} \times 2\,\text{h} = 2.2\,\text{kWh}$$

Result:
- Power of the motor $= 1100\,\text{W}$
- Energy consumed in 2 h $= 7.92 \times 10^6\,\text{J}$ (or $2.2\,\text{kWh}$)

Correct option: (d) 25

Justification:

When the wire of resistance $R$ is cut into 5 equal parts, the resistance of each part is:
$$r = \frac{R}{5}$$

When these 5 equal resistors are connected in parallel, the equivalent resistance $R'$ is:
$$\frac{1}{R'} = \frac{1}{r} + \frac{1}{r} + \frac{1}{r} + \frac{1}{r} + \frac{1}{r} = \frac{5}{r}$$
$$R' = \frac{r}{5} = \frac{R/5}{5} = \frac{R}{25}$$

Therefore:
$$\frac{R}{R'} = \frac{R}{R/25} = 25$$

Correct option: (b) $IR^2$

Justification:

Electrical power is given by $P = VI = I^2R = \dfrac{V^2}{R}$.

The expression $IR^2$ has units of $\text{A} \times \Omega^2$, which does not simplify to watts (W). Hence $IR^2$ does not represent electrical power.

Correct option: (d) 25 W

Justification:

The resistance of the bulb (assumed constant):
$$R = \frac{V^2}{P} = \frac{(220)^2}{100} = 484\,\Omega$$

Power consumed at 110 V:
$$P' = \frac{V'^2}{R} = \frac{(110)^2}{484} = \frac{12100}{484} = 25\,\text{W}$$

Correct option: (c) 1:4

Justification:

Let the resistance of each wire be $R$.

- In series: $R_s = R + R = 2R$
- In parallel: $R_p = \dfrac{R \times R}{R + R} = \dfrac{R}{2}$

Heat produced $= \dfrac{V^2}{R_{eq}} \times t$ (at same potential difference $V$ and time $t$):

$$H_s = \frac{V^2}{2R} \times t, \quad H_p = \frac{V^2}{R/2} \times t = \frac{2V^2}{R} \times t$$

$$\frac{H_s}{H_p} = \frac{V^2/(2R)}{2V^2/R} = \frac{1}{4}$$

So the ratio is $1:4$.

A voltmeter is always connected in parallel across the two points between which the potential difference is to be measured.

Reason: A voltmeter has very high resistance so that it draws negligible current from the circuit and does not disturb the original potential difference across those two points.

Given:
- Diameter, $d = 0.5\,\text{mm} = 0.5 \times 10^{-3}\,\text{m}$
- Radius, $r = 0.25 \times 10^{-3}\,\text{m}$
- Resistivity, $\rho = 1.6 \times 10^{-8}\,\Omega\text{m}$
- Resistance, $R = 10\,\Omega$

Step 1: Find the area of cross-section
$$A = \pi r^2 = \pi \times (0.25 \times 10^{-3})^2 = \pi \times 6.25 \times 10^{-8}\,\text{m}^2$$
$$A = 1.964 \times 10^{-7}\,\text{m}^2$$

Step 2: Find the length using $R = \rho \dfrac{l}{A}$
$$l = \frac{R \times A}{\rho} = \frac{10 \times 1.964 \times 10^{-7}}{1.6 \times 10^{-8}}$$
$$l = \frac{1.964 \times 10^{-6}}{1.6 \times 10^{-8}} = 122.7\,\text{m} \approx 122.7\,\text{m}$$

Step 3: Effect of doubling the diameter

If diameter is doubled, new diameter $d' = 2d$, so new radius $r' = 2r$.
$$A' = \pi (2r)^2 = 4\pi r^2 = 4A$$

New resistance:
$$R' = \rho \frac{l}{A'} = \rho \frac{l}{4A} = \frac{R}{4} = \frac{10}{4} = 2.5\,\Omega$$

Result:
- Length of wire required $= 122.7\,\text{m}$
- When diameter is doubled, resistance becomes $2.5\,\Omega$ (i.e., it reduces to one-fourth of the original value).

Given data:

| $I$ (A) | 0.5 | 1.0 | 2.0 | 3.0 | 4.0 |
|---------|-----|-----|-----|------|------|
| $V$ (V) | 1.6 | 3.4 | 6.7 | 10.2 | 13.2 |

Graph: Plot $V$ on the y-axis and $I$ on the x-axis. The points approximately lie on a straight line passing through (or near) the origin, confirming Ohm's law.

Calculating Resistance:

The resistance is the slope of the $V$–$I$ graph:
$$R = \frac{\Delta V}{\Delta I}$$

Using two well-separated points, e.g., $(I = 0.5\,\text{A},\, V = 1.6\,\text{V})$ and $(I = 4.0\,\text{A},\, V = 13.2\,\text{V})$:
$$R = \frac{13.2 - 1.6}{4.0 - 0.5} = \frac{11.6}{3.5} \approx 3.31\,\Omega$$

We can also verify with individual readings:
- $R = 1.6/0.5 = 3.2\,\Omega$
- $R = 3.4/1.0 = 3.4\,\Omega$
- $R = 6.7/2.0 = 3.35\,\Omega$
- $R = 10.2/3.0 = 3.4\,\Omega$
- $R = 13.2/4.0 = 3.3\,\Omega$

Average (from graph slope) $\approx 3.4\,\Omega$

The resistance of the resistor is approximately $\mathbf{3.4\,\Omega}$.

Given:
- Voltage, $V = 12\,\text{V}$
- Current, $I = 2.5\,\text{mA} = 2.5 \times 10^{-3}\,\text{A}$

Using Ohm's Law:
$$R = \frac{V}{I} = \frac{12}{2.5 \times 10^{-3}} = \frac{12}{0.0025} = 4800\,\Omega$$

The resistance of the resistor is $4800\,\Omega$ (or $4.8\,\text{k}\Omega$).

Given:
- EMF of battery, $V = 9\,\text{V}$
- Resistors in series: $0.2\,\Omega,\; 0.3\,\Omega,\; 0.4\,\Omega,\; 0.5\,\Omega,\; 12\,\Omega$

Step 1: Find total resistance in series
$$R_{total} = 0.2 + 0.3 + 0.4 + 0.5 + 12 = 13.4\,\Omega$$

Step 2: Find current using Ohm's Law

In a series circuit, the same current flows through all resistors:
$$I = \frac{V}{R_{total}} = \frac{9}{13.4} \approx 0.671\,\text{A}$$

The current through the $12\,\Omega$ resistor is approximately $0.67\,\text{A}$.

Given:
- Each resistor, $R = 176\,\Omega$
- Total current, $I = 5\,\text{A}$
- Voltage, $V = 220\,\text{V}$

Step 1: Find the required equivalent resistance
$$R_{eq} = \frac{V}{I} = \frac{220}{5} = 44\,\Omega$$

Step 2: Find number of resistors in parallel

For $n$ resistors of resistance $R$ each in parallel:
$$R_{eq} = \frac{R}{n}$$
$$n = \frac{R}{R_{eq}} = \frac{176}{44} = 4$$

4 resistors of $176\,\Omega$ connected in parallel are required.

Each resistor has resistance $R = 6\,\Omega$.

(i) To get $9\,\Omega$:

Connect two resistors in parallel, and then connect the third resistor in series with this parallel combination.

- Parallel combination of two $6\,\Omega$ resistors:
$$R_p = \frac{6 \times 6}{6 + 6} = \frac{36}{12} = 3\,\Omega$$

- Series with the third resistor:
$$R_{total} = 3 + 6 = 9\,\Omega \checkmark$$

(ii) To get $4\,\Omega$:

Connect two resistors in series, and then connect the third resistor in parallel with this series combination.

- Series combination of two $6\,\Omega$ resistors:
$$R_s = 6 + 6 = 12\,\Omega$$

- Parallel with the third resistor:
$$R_{total} = \frac{12 \times 6}{12 + 6} = \frac{72}{18} = 4\,\Omega \checkmark$$

Given:
- Voltage of supply, $V = 220\,\text{V}$
- Power rating of each bulb, $P = 10\,\text{W}$
- Maximum allowable current, $I_{max} = 5\,\text{A}$

Step 1: Find current drawn by each bulb
$$I_{bulb} = \frac{P}{V} = \frac{10}{220} = \frac{1}{22}\,\text{A}$$

Step 2: Find number of bulbs

In parallel, total current = $n \times I_{bulb}$
$$n = \frac{I_{max}}{I_{bulb}} = \frac{5}{1/22} = 5 \times 22 = 110$$

110 lamps can be connected in parallel.

Given:
- Voltage, $V = 220\,\text{V}$
- Resistance of each coil: $R_A = R_B = 24\,\Omega$

Case 1: Used separately (only one coil)
$$I = \frac{V}{R} = \frac{220}{24} \approx 9.17\,\text{A}$$

Case 2: Connected in series
$$R_s = R_A + R_B = 24 + 24 = 48\,\Omega$$
$$I_s = \frac{V}{R_s} = \frac{220}{48} \approx 4.58\,\text{A}$$

Case 3: Connected in parallel
$$R_p = \frac{R_A \times R_B}{R_A + R_B} = \frac{24 \times 24}{24 + 24} = \frac{576}{48} = 12\,\Omega$$
$$I_p = \frac{V}{R_p} = \frac{220}{12} \approx 18.33\,\text{A}$$

Summary:
- Separately: $\approx 9.17\,\text{A}$
- In series: $\approx 4.58\,\text{A}$
- In parallel: $\approx 18.33\,\text{A}$

Case (i): 6 V battery in series with 1 Ω and 2 Ω

Total resistance:
$$R_{total} = 1 + 2 = 3\,\Omega$$

Current through circuit (same through both resistors in series):
$$I = \frac{V}{R_{total}} = \frac{6}{3} = 2\,\text{A}$$

Power used in $2\,\Omega$ resistor:
$$P_1 = I^2 \times R = (2)^2 \times 2 = 4 \times 2 = 8\,\text{W}$$

Case (ii): 4 V battery in parallel with 12 Ω and 2 Ω

In a parallel circuit, the voltage across each resistor equals the battery voltage.

Voltage across $2\,\Omega$ resistor $= 4\,\text{V}$

Power used in $2\,\Omega$ resistor:
$$P_2 = \frac{V^2}{R} = \frac{(4)^2}{2} = \frac{16}{2} = 8\,\text{W}$$

Comparison:
$$\frac{P_1}{P_2} = \frac{8}{8} = 1$$

The power used in the $2\,\Omega$ resistor is equal (8 W each) in both circuits.

Given:
- Lamp 1: $P_1 = 100\,\text{W}$ at $220\,\text{V}$
- Lamp 2: $P_2 = 60\,\text{W}$ at $220\,\text{V}$
- Supply voltage, $V = 220\,\text{V}$

Current drawn by each lamp:
$$I_1 = \frac{P_1}{V} = \frac{100}{220} = \frac{5}{11}\,\text{A}$$
$$I_2 = \frac{P_2}{V} = \frac{60}{220} = \frac{3}{11}\,\text{A}$$

Total current drawn from the line (parallel connection):
$$I = I_1 + I_2 = \frac{5}{11} + \frac{3}{11} = \frac{8}{11} \approx 0.727\,\text{A}$$

The current drawn from the line is approximately $0.73\,\text{A}$.

Given:
- Resistance, $R = 44\,\Omega$
- Current, $I = 5\,\text{A}$
- Time, $t = 2\,\text{h}$ (not needed for rate)

The rate at which heat is developed = Power dissipated:
$$P = I^2 R = (5)^2 \times 44 = 25 \times 44 = 1100\,\text{W}$$

The rate at which heat is developed in the heater is $1100\,\text{W}$ (or $1.1\,\text{kW}$).

(a) Tungsten for filament of electric lamps:

Tungsten is used almost exclusively for filaments because:
1. It has a very high melting point ($3380°\text{C}$), so it can be heated to very high temperatures (around $2500°\text{C}$) without melting.
2. It has high resistivity, which allows it to produce a large amount of heat and light.
3. It does not oxidise readily at high temperatures (especially in an inert gas atmosphere inside the bulb).

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(b) Alloys for heating devices instead of pure metals:

Alloys are preferred over pure metals in heating devices because:
1. Alloys have higher resistivity than pure metals, so they produce more heat for the same current.
2. Alloys do not oxidise (burn) easily at high temperatures, making them more durable.
3. They have a high melting point, so they can withstand the high temperatures generated during operation.

*(Example: Nichrome is used in toasters and electric irons.)*

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(c) Series arrangement not used for domestic circuits:

Series arrangement is not used in domestic circuits because:
1. In series, if one appliance fails or is switched off, the circuit breaks and all other appliances stop working.
2. All appliances in series get the same current, but different appliances require different currents for proper operation.
3. The voltage gets divided among all appliances, so each appliance does not get the required voltage for proper functioning.
4. We cannot independently control individual appliances.

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(d) Resistance of a wire and its area of cross-section:

The resistance of a wire is given by:
$$R = \rho \frac{l}{A}$$
where $A$ is the area of cross-section.

Resistance is inversely proportional to the area of cross-section:
$$R \propto \frac{1}{A}$$

This means: as the area of cross-section increases, the resistance decreases, and vice versa. A thicker wire offers less resistance because more electrons can flow through it simultaneously.

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(e) Copper and aluminium for electricity transmission:

Copper and aluminium are used for electricity transmission because:
1. Both have very low resistivity (copper: $1.7 \times 10^{-8}\,\Omega\text{m}$; aluminium: $2.8 \times 10^{-8}\,\Omega\text{m}$), so they are excellent conductors and energy loss during transmission is minimal.
2. They are good conductors of electricity, allowing large currents to flow with minimal heating.
3. They are ductile (can be drawn into thin wires) and relatively inexpensive and abundant.
4. Aluminium is also lightweight, making it suitable for overhead transmission lines.