Question

A current of 12 A flows in the element of an electric fire of resistance 25 Ω. Determine the power dissipated by the element. If the fire is on for five hours every day, calculate for a one-week period (a) the energy used, and (b) cost of using the fire if electricity cost ₱13.50 per kWh

Answer 1

Given:

$I=12A;$ $R=25\Omega;$

$td=5h;$ -time for day; $tw=7\times5h=35h.$ - time for week.

Solution:

$P=I^{2}R$ -power dissipated by the element;

$P=(12A)^{2}\times25\Omega=3600W.$

a) $E=Ptw$ - energy used for one-week;

$E=3600W\times35h=126kWh.$

b) $Ct=E\times13.5\times(\frac{₱}{kWh})$ - cost of electric energy per week;

$Ct=126kWh\times13.5\times(\frac{₱}{kWh})=1701₱.$

Answer:

$P=3600W;$

a) $E=126kWh;$

b) $Ct=1701₱.$