Question

a electric heater of resistance 10w draws 15A current from the service mains for 2hours. calculate (1)the heat developed in the heater and (2) the power of the heater​

Answer 1

Given :-

Resistance (R)= 10 ohm

Current (I) = 15 A

Time (t)= 2 hours = 2×60×60

Time (t) = 7200 seconds

To find :-

• Heat developed in the heater

• The power of heater

• First find the heat

$\underline{\small{\boxed{\sf{\dag\ \ Heat(H)= [current (I)]^2\times Resistance(R)\times Time (t)}}}}$

• By putting the values

$:\implies\sf H= I^2Rt\\ \\ :\implies\sf H= (15)^2\times 10\times 7200\\ \\ :\implies\sf H= 225\times 72000\\ \\ :\implies\sf H=16200000\ joules \\ \\ \bullet\sf \ 1 \ joules = \dfrac{1}{1000} \ kilojoules\\ \\ :\implies\sf H= 16200\ kilojoules$

$\underline{\small{\boxed{\sf{\dag\ \ Heat \ produced\ by \ heater = 16,200\ kilojoules}}}}$

• Now calculate the power

$\underline{\small{\boxed{\sf{\dag\ \ Power (P)= [Current (I)]^2\times Resistance (R)}}}}$

• We have ,

• Resistance (R) = 10 ohm

• Current (I) = 15 A

• Power (P)=?

$:\implies \sf P= I^2R\\ \\ :\implies\sf P= (15)^2\times 10\\ \\ :\implies\sf P= 225\times 10\\ \\ :\implies\sf P= 2250\ watts$

$\underline{\boxed{\sf{\dag\ \ Power \ of \ heater = 2250 \ watts }}}$

Answer 2

Given

● Resistance of heater = 10 ohm.

● Current in heater = 15A

● Time = 2 hours.

To find

● Heat developed in heater

● Power of the heater

Solution

1) Here resistance (r) = 10Ω

◕ Current (I) = 15A

◕ Time (t) = 2 hours.

⇒ Time (t) = (2 × 3600) s

⇒ Time (t) = 7200 s

We know that,

$\dag \: \: \underline{\boxed{\sf\blue{Heat(H) = Current(I)^2 \times Resistance (R) \times Time(t) }}}\\\\ \longmapsto\sf Heat \: developed(H) = \bigg\lgroup (15)^2 \times 10 \times 7200 \bigg\rgroup \\\\ \longmapsto\sf Heat \: developed (H) = \bigg\lgroup 225 \times 10 \times 7200 \bigg\rgroup \\\\ \longmapsto\sf Heat \: developed = 16200000 J \\\\ \longmapsto\underline{\underline{\textsf{\textbf{Heat developed (H) = 16,200 kJ }}}} \\\\\\ \therefore\underline{\boxed{\textsf{Heat developed in heater = {\textbf{16200 kJ }}}}}$

$\rule{200}{1}$

(2) We know that,

$\dag\: \: \underline{\boxed{\sf\green{Power(P) = Current (I)^2 \times Resistance (R) }}}\\\\ \longmapsto\sf Power_{heater} = \bigg\lgroup (15)^2 \times 10 \bigg\rgroup \\\\ \longmapsto\sf Power_{heater} = \bigg\lgroup 225 \times 10 \bigg\rgroup \\\\ \longmapsto\sf Power_{heater} = 2250 \: Watts \\\\\longmapsto\underline{\underline{\textsf{\textbf{ Power in heater = 2.25\: Kilowatts }}}}\\\\\\ \therefore\underline{\boxed{\textsf{Power in heater = {\textbf{2.25 Kilowatts }}}}}$