Question

An electric heater of resistance 8Ω draws 15A from the service mains for 2 hours. Calculate the rate at which heat is developed in the heater.​

Answer 1

Given

• I = 15A
• R = 8Ω
• t = 2 hours

To find

• The rate at which heat is developed, i.e., electric power.

Solution

$\sf\pink{⟶}$ In this question, we have to find the electric power (P).

★ We know that

$\large{\underline{\boxed{\tt{P = I^2R}}}}$

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$\tt:\implies\: \: \: \: \: \: \: \: {P = (15)^2 \times 8}$

⠀

$\tt:\implies\: \: \: \: \: \: \: \: {P = 225 \times 8}$

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$\tt:\implies\: \: \: \: \: \: \: \: {P = 1800 W}$

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$\therefore$ Rate at which heat is developed is 1800 J/s.

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$\: \: \: \: \: \: \: \: \: \: \boxed{\bigstar{\bf{More\: to\: know\: {\bigstar}}}}$

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$\longmapsto$ Electric power, P is the rate at which electric energy is consumed in an electric circuit.

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$\longmapsto$ Formulae to calculate electric power:

$\large\dag{\underline{\boxed{\orange{P = VI}}}}$

$\large\dag{\underline{\boxed{\orange{P = I^2R}}}}$

$\large\dag{\underline{\boxed{\orange{P = \dfrac{V^2}{R}}}}}$

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Answer 2

ɢɪᴠᴇɴ:-

• Resistane (R) = 8 Ω
• Electric current (I) = 15 A
• Time (T) = 2 h = 2 × 60 × 60 = 7200 sec

ᴛᴏ ꜰɪɴᴅ:-

• Electric Power (P)

ꜱᴏʟᴜᴛɪᴏɴ:-

As we know that,

$\red{\boxed{\gray{\sf{H = I^2 R T}}}}$

$\begin{lgathered}\begin{lgathered}\begin{lgathered}:\implies\frak {H\: = (15)^2 \times 8 \times 7200} \\ \\\end{lgathered}\end{lgathered}\end{lgathered}$

$\begin{lgathered}\begin{lgathered}\begin{lgathered}:\implies\frak {H \: = 225 \times 57,600} \\ \\\end{lgathered}\end{lgathered}\end{lgathered}$

$\begin{lgathered}\begin{lgathered}\begin{lgathered}:\implies\frak {H = 12,960,000} \\ \\\end{lgathered}\end{lgathered}\end{lgathered}$

$\begin{lgathered}\begin{lgathered}\begin{lgathered}:\implies\frak\green{H \: = \: 12,960,000 \: J} \\ \\\end{lgathered}\end{lgathered}\end{lgathered}$

$\tt \pink{\boxed{\purple{\sf{Power = \frac{Heat}{Time}}}}}$

$\begin{lgathered}\begin{lgathered}\begin{lgathered}:\implies\frak {Power = \frac{12960000}{7200}} \\ \\\end{lgathered}\end{lgathered}\end{lgathered}$

$\begin{lgathered}\begin{lgathered}\begin{lgathered}:\implies\frak {Power = 1800} \\ \\\end{lgathered}\end{lgathered}\end{lgathered}$

$\begin{lgathered}\begin{lgathered}\begin{lgathered}:\implies\frak\pink{Power \: = \: 1800 \: Js^{-1}} \\ \\\end{lgathered}\end{lgathered}\end{lgathered}$

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