A room `AC` run for `5` hour at a voltage of `220V` The wiring of the room constant of `Cu` of `1mm` ratio and a length of `10 m` consumption per day is `10` commerclal unit What fraction of it goes in the joule heated in wire? What would happen if the wiring is made of aluminum of the same distances? `[rho_(cu) = 1.7 xx 10^(-8) Omega,rho_(A1) = 2.7 xx 10^(-8) Omega m]`

Answer

To solve the given problem step by step, we will follow the outlined process:

### Step 1: Calculate the Power Consumption
The power consumption per day is given as 10 commercial units. One commercial unit is equivalent to 1 kilowatt-hour (kWh). Therefore, the total energy consumed in a day is:
\[ \text{Energy} = 10 \text{ kWh} = 10 \times 1000 \text{ Wh} = 10000 \text{ Wh} \]
Since the AC runs for 5 hours, we can find the power consumed:
\[ \text{Power} = \frac{\text{Energy}}{\text{Time}} = \frac{10000 \text{ Wh}}{5 \text{ h}} = 2000 \text{ W} = 2000 \text{ J/s} \]

### Step 2: Calculate the Resistance of the Copper Wire
The resistance \( R \) of the wire can be calculated using the formula:
\[ R = \frac{\rho L}{A} \]
Where:
- \( \rho \) (resistivity of copper) = \( 1.7 \times 10^{-8} \, \Omega \, \text{m} \)
- \( L \) (length of the wire) = 10 m
- \( A \) (cross-sectional area) = \( \pi r^2 \)

The radius \( r \) is given as 1 mm, which is \( 1 \times 10^{-3} \, \text{m} \). Thus,
\[ A = \pi (1 \times 10^{-3})^2 = \pi \times 10^{-6} \approx 3.14 \times 10^{-6} \, \text{m}^2 \]
Now substituting the values into the resistance formula:
\[ R = \frac{1.7 \times 10^{-8} \times 10}{3.14 \times 10^{-6}} \approx 0.539 \, \Omega \]

### Step 3: Calculate the Current
Using the power formula \( P = VI \), we can rearrange it to find the current \( I \):
\[ I = \frac{P}{V} = \frac{2000 \text{ W}}{220 \text{ V}} \approx 9.09 \text{ A} \]

### Step 4: Calculate the Power Loss in the Copper Wire
The power loss due to Joule heating in the wire is given by:
\[ P_{\text{loss}} = I^2 R \]
Substituting the values:
\[ P_{\text{loss}} = (9.09)^2 \times 0.539 \approx 4.38 \text{ W} \]

### Step 5: Calculate the Fraction of Power Loss
To find the fraction of power loss in the wire:
\[ \text{Fraction} = \frac{P_{\text{loss}}}{P_{\text{total}}} = \frac{4.38}{2000} \times 100 \approx 0.219\% \]

### Step 6: Calculate Power Loss if the Wiring is Made of Aluminum
Now, we will repeat the process for aluminum. The resistivity of aluminum is given as \( 2.7 \times 10^{-8} \, \Omega \, \text{m} \).

#### Step 6.1: Calculate the Resistance of the Aluminum Wire
\[ R_{\text{Al}} = \frac{\rho_{\text{Al}} L}{A} = \frac{2.7 \times 10^{-8} \times 10}{3.14 \times 10^{-6}} \approx 0.859 \, \Omega \]

#### Step 6.2: Calculate the Power Loss in the Aluminum Wire
\[ P_{\text{loss, Al}} = I^2 R_{\text{Al}} = (9.09)^2 \times 0.859 \approx 6.95 \text{ W} \]

#### Step 6.3: Calculate the Fraction of Power Loss for Aluminum
\[ \text{Fraction}_{\text{Al}} = \frac{P_{\text{loss, Al}}}{P_{\text{total}}} = \frac{6.95}{2000} \times 100 \approx 0.345\% \]

### Final Results
- Fraction of power loss in copper wire: **0.219%**
- Fraction of power loss in aluminum wire: **0.345%**