Question

Calculate the heat developed when 2A current flows for 3 minutes through an electric heater of resistance 920 ohm​

Answer 1

$\sf Given \begin {cases} & \sf { Current \: = \: 20\: A } \\ & \sf{ Resistance = \: 920 \ohm} \\ &\sf { time = \: 3 min = 180 sec } \end {cases}$

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$\sf To \: find \begin {cases} & \sf { Heat\: = \: ?? } \\ \end {cases}$

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Using Joule's equation of electrical heating :-

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$\star\;{\boxed{\sf{\pink{Heat ( H ) = Current ( I ) ^2 \times Resistance ( R ) \times Time ( T) }}}}$

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$:\implies\sf H = 2 \times 2 \times 920 \times 180$

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$:\implies\sf H = 4 \times 920 \times 180$

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$:\implies \sf H = 3680 \times 180$

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$:\implies \sf H = 662400 Joules$

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$:\implies{\underline{\boxed{\frak{\purple{ Heat \: = 662400\: joules\: }}}}}\;\bigstar$

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Answer 2

Given : The amount of Current flows is 2A( Ampere ) , Resistance is the 920 ohm & Total time taken is 3 min .

Exigency To Find : The amount of Heat Developed.

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⠀⠀⠀⠀⠀⠀⠀Given that ,

⠀⠀⠀⠀⠀▪︎⠀⠀The amount of Current [ I ] flows is 2A( Ampere ).

⠀⠀⠀⠀⠀▪︎⠀⠀Resistance [ R ] is 920 $\Omega$ .

⠀⠀⠀⠀⠀▪︎⠀⠀Total Time taken [ T ] is 3 min .

⠀⠀⠀⠀⠀⠀Converting Time taken from minutes to second :

$\qquad:\implies \sf Total \:Time \:Taken\:\:(\ T\ ) \:=\: 3\:min \:$

$\qquad:\implies \sf Total \:Time \:Taken\:\:(\ T\ ) \:=\: 3\times 60\:second \:\qquad \bigg\lgroup \sf{ 1\: minute \:=\:60\:seconds\:}\bigg\rgroup \:\:$

$\qquad:\implies \bf Total \:Time \:Taken\:\:(\ T\ ) \:=\: 180\:seconds \:$

$\qquad :\implies \pmb{\underline{\purple{\: Total \:Time \:Taken\:\:(\ T\ ) \:=\: 180\:seconds\: }} }\:\bigstar$

⠀⠀⠀⠀⠀⠀Now ,

$\dag\:\:\sf{ As,\:We\:know\:that\::}\\\\ \qquad\bigstar \:\:\bf \:By \:Joule's \:Equation\:of\:Electrical\:Heating\::$

$\qquad \dag\:\:\bigg\lgroup \sf{\: H \:=\:(\ I\ )^2 \:\times\: R \:\times T \:\: }\bigg\rgroup$

⠀⠀⠀⠀⠀⠀Here , I is the amount of Current , R is the Resistance & T is the Total time taken .

$\qquad:\implies \sf H \:=\:(\ I\ )^2 \:\times\: R \:\times T \:\:$

$\qquad:\implies \sf H \:=\:(\ 2 \ )^2 \:\times\: 920 \:\times 180 \:\:$

$\qquad:\implies \sf H \:=\:4 \:\times\: 920 \:\times 180 \:\:$

$\qquad:\implies \sf H \:=\:4 \:\times\: 165600 \:\:$

$\qquad:\implies \sf H \:=\:662400 \:\:$

$\qquad :\implies \pmb{\underline{\purple{\: Heat\:\:(\ H \ ) \:=\: 662400\:Joules\: }} }\:\bigstar$

⠀⠀⠀⠀⠀▪︎⠀⠀ Here , H denotes Heat which is 662400 Joules

Therefore,

⠀⠀⠀⠀⠀$\therefore {\underline{ \mathrm {\:The \:Heat\:Developed \:is\:\bf{662400\;Joules \:}}.}}$

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