Question

A room has AC run hours a day at a voltage of $220$V. The wirng of the room consists of Cu of $1$mm radius and a length of $10$m. Power consumption per day is $10$ commercial units . The fraction of it goes in the joule heating in wires is: ${\rho }_{cu}=1.7\times {10}^{-8}\Omega m$

• A. $0.32\%$
• B. $0.2\%$
• C. $2\%$
• D. $3.2\%$

Answer 1

The correct option is A $0.32\%$

Power consumption $=\frac{10units}{5hour}=\frac{2units}{hour}$

$=2kw$

$=2000J/S$

So, $I=\frac{P}{V}$

$=\frac{2000}{220}$

$=9.09A$

Power loss in copper wire $=I^R$

$=I^2\frac{PL}{A}$

Power loss in copper wire $=9.09\times 9.09\times \frac{1.7\times {10}^{-8}\times 10}{17\times {\left({10}^{-3}\right)}^2}$

$=4.47J$

$=\frac{4.47}{2000}\times 100\%$

$=0.22\%$

Power loss in aluminium wire $=I^{2}R=I^2\frac{PL}{A}$

Power loss in copper wire $=9.09\times 9.09\times \frac{2.7\times {10}^{-8}\times 10}{17\times {\left({10}^{-3}\right)}^2}$

$=7.1J$

$=\frac{7.1}{2000}\times 100\%$

$=0.355\%$

Hence, the answer is $0.355\%.$