Vardaan Learning Institute

Potential difference (Voltage) is defined as the work done to move a unit charge. $V = W/Q$.

$1 \text{ kWh} = 1000 \text{ W} \times 3600 \text{ s} = 3.6 \times 10^6 \text{ Joules}$.

In a series combination, the current remains the same through all components.

This works for all cases. While (a) and (c) are also valid forms derived from Ohm's law, (b) is the standard mathematical expression of Joule's Law of Heating. If "All of these" is an option and they are all algebraically equivalent, then (d) is correct. Given the options, (d) is the most comprehensive answer.

A fuse melts when excessive current generates heat beyond its melting point.

$H \propto I^2$. If $I$ becomes $I/2$, then $H' \propto (I/2)^2 = I^2/4$. Heat becomes 1/4th.

Tungsten is used because of its very high melting point ($\sim 3380^\circ$C) so it doesn't melt at high temperatures, and high resistivity to generate more heat/light.

Power is the rate of consumption of energy. $P = E/t$.

Analysis of Circuit Diagram:
The circuit has a top branch with $R_1 = 5\Omega$ and a bottom branch with $R_2 = 4\Omega$ in series with $R_3 = 6\Omega$. Total resistance of bottom branch $R_{bottom} = 4+6 = 10\Omega$.
Since branches are parallel, Voltage (V) is the same.
Heat/sec (Power) in top branch: $P_1 = V^2 / 5 = 10 \implies V^2 = 50$.
Current in bottom branch: $I_{bottom} = V / R_{bottom} = \sqrt{50} / 10$.
Heat/sec in $4\Omega$ resistor: $P_2 = I_{bottom}^2 \times 4$.
$P_2 = (\frac{\sqrt{50}}{10})^2 \times 4 = \frac{50}{100} \times 4 = 0.5 \times 4 = 2 \text{ cal/sec}$.

Analysis of Graph:
The graph plots Heat ($H$) vs Current Squared ($I^2$).
From Joule's Law: $H = I^2 R t$.
Comparing to equation of a line $y = mx$, here $y=H$ and $x=I^2$.
The slope $m = H/I^2 = R t$.

Heat produced $H = I^2Rt$. Since current is same in series, $H \propto R$. The element has very high resistance (high heat, glows), while wires have negligible resistance (negligible heat).

Assertion is False: A fuse is always connected in series, not parallel. If it were in parallel, it wouldn't break the main circuit when it melts.

In parallel, $V$ is constant. Power $P = V^2/R \implies P \propto 1/R$. The 200W bulb has higher power, meaning lower resistance, drawing more current and glowing brighter.

1. High Resistivity: Alloys (like Nichrome) have higher resistivity than pure metals, producing more heat for the same current.
2. No Oxidation: Alloys do not oxidize (burn) easily at high temperatures compared to pure metals, ensuring durability.

Resistance of bulb: $R = V^2/P = 220^2/100 = 484 \Omega$.
Power at 110V: $P' = V'^2/R = 110^2 / 484 = 12100 / 484 = 25 \text{ W}$.
Alternatively: Voltage is halved. Since $P \propto V^2$, Power becomes $(1/2)^2 = 1/4$th. $100/4 = 25$W.

Statement: The heat produced in a resistor is directly proportional to:
(i) Square of current ($I^2$) for a given resistance.
(ii) Resistance ($R$) for a given current.
(iii) Time ($t$) for which the current flows.
Formula: $H = I^2 R t$.

Given: Charge $Q = 96000$ C, $V = 50$ V.
Heat generated = Work Done.
$H = V \times Q = 50 \times 96000 = 48,00,000 \text{ J} = 4.8 \times 10^6 \text{ J}$.

Tungsten is used because:
1. It has a very high melting point ($3380^\circ$C).
2. It retains strength at high temperatures.
3. It emits light at high temperatures (incandescence).

Power adds up in parallel: $P_{total} = P_1 + P_2 = 100 + 60 = 160$ W.
Current $I = P/V = 160 / 220 = 16/22 \approx 0.73 \text{ A}$.
Alternatively, calculate individual currents: $I_1 = 100/220 \approx 0.45$A, $I_2 = 60/220 \approx 0.27$A. Total $\approx 0.72-0.73$A.

TV Set: $P=250W = 0.25kW$, $t=1hr$. $E = P \times t = 0.25 \times 1 = 0.25 \text{ kWh}$.
Toaster: $P=1200W = 1.2kW$, $t=10min = 1/6 hr$. $E = 1.2 \times 1/6 = 0.2 \text{ kWh}$.
Conclusion: The TV set uses more energy ($0.25 > 0.20$).

Rate of heat development is Power.
$P = I^2 R = (15)^2 \times 8 = 225 \times 8 = 1800 \text{ J/s (Watts)}$.

Function: A safety device that breaks the circuit if current exceeds a safe limit, preventing fire or damage.
Characteristics:
1. Made of an alloy (lead and tin) with a low melting point so it melts quickly upon overheating.
2. Placed in series with the live wire.
3. High resistivity compared to mains copper wire (ensures heat is generated there).

Max Current $I = 5$A, Voltage $V = 220$V.
Max Power allowed $P_{max} = V \times I = 220 \times 5 = 1100$ W.
Power of one bulb $P_{bulb} = 10$ W.
Number of bulbs $n = P_{max} / P_{bulb} = 1100 / 10 = 110$.
110 lamps can be connected.

Supply V = 220V.
(i) Separately: $R = 24\Omega$. $I = V/R = 220/24 \approx 9.16$ A.
(ii) Series: $R_s = 24+24 = 48\Omega$. $I = 220/48 \approx 4.58$ A.
(iii) Parallel: $R_p = 24/2 = 12\Omega$. $I = 220/12 \approx 18.33$ A.

(a) Electric power is the rate at which electrical energy is consumed. $P = VI = V(V/R) = V^2/R$.
(b) It means the fuse wire will melt and break the circuit if the current flowing through it exceeds 2 Amperes.
(c) Current drawn by iron: $I = P/V = 1000 W / 220 V \approx 4.54$ A.
Since the current is 4.54 A, the fuse must be rated slightly higher than this. A 5 A fuse is suitable. (1A and 3A will melt immediately).

(i) Iron. Iron has higher resistivity ($\rho$) than copper. Since $R = \rho L/A$ and dimensions are same, Iron has higher resistance.
(ii) Copper (One by one - Parallel/Same V): $H = V^2t/R$. Heat is inversely proportional to R. Copper has lower resistance, so it produces more heat.
(iii) Iron (Series - Same I): $H = I^2Rt$. Heat is directly proportional to R. Iron has higher resistance, so it produces more heat.
(iv) Justification provided in steps above.

Energy per day:
(i) Fridge: $400W \times 10h = 4000$ Wh.
(ii) Fans: $2 \times 80W \times 12h = 1920$ Wh.
(iii) Tubes: $6 \times 18W \times 6h = 648$ Wh.
Total Energy/day = $4000 + 1920 + 648 = 6568 \text{ Wh} = 6.568 \text{ kWh}$.
For June (30 days): $6.568 \times 30 = 197.04 \text{ kWh (units)}$.
Cost: $197.04 \times 3.00 =$ ₹ $591.12$.

Analysis of Circuit Diagram: Supply 6V. Resistors $5\Omega, 10\Omega, 30\Omega$ in parallel.
(a) Current through each: Since parallel, $V=6V$ for all.
- $I_1 = 6/5 = 1.2$ A.
- $I_2 = 6/10 = 0.6$ A.
- $I_3 = 6/30 = 0.2$ A.
(b) Total Current: $I = 1.2 + 0.6 + 0.2 = 2.0$ A.
(c) Effective Resistance: $R_{eff} = V/I = 6/2 = 3\Omega$. (Or use $1/R$ formula).
(d) Heat in $5\Omega$ (10s): $H = V^2/R \times t = 6^2/5 \times 10 = 36/5 \times 10 = 36 \times 2 = 72$ Joules.

(a) To minimize power loss ($I^2R$) in transmission lines. High voltage implies low current for the same power ($P=VI$), significantly reducing heat loss.
(b) $1 \text{ kWh} = 3.6 \times 10^6 \text{ Joules}$.
(c) Energy $E = P \times t = 1.5 \text{ kW} \times 10 \text{ h} = 15 \text{ kWh}$.
In Joules: $15 \times 3.6 \times 10^6 = 54 \times 10^6 = 5.4 \times 10^7 \text{ J}$.

(i) Nichrome (an alloy of Nickel, Chromium, Manganese, and Iron).
(ii) $H = I^2 R t = 2^2 \times 100 \times 10 = 4 \times 1000 = 4000 \text{ J}$.
(iii) The heating element has high resistance, producing large amounts of heat ($H \propto R$), causing it to become red hot. The connecting cord (copper) has very low resistance, producing negligible heat, so it doesn't glow.
(iv) To increase heat ($P=V^2/R$), we must decrease the resistance. Cutting the coil shorter reduces its length and thus its resistance, increasing power output (though this may reduce its life).