Question

calculate the heat produced in calorie when a current of 0.1 ampere is passed through a wire of resistance 41.8 ohms for 5 minut​

Answer 1

I = 0.1A

R = 41.8 ohm

t = 5min = 5×60 sec

H = I²Rt

H = (0.1)²×41.8×5×60 = 125.4J

1 Cal = 4.18J

1J = 1/4.18 Cal

125.5J = 125.4/4.18 = 30Cal

Hope this is helpful to you

Pls mark as brainliest

Follow me and I'll follow you back

Answer 2

$\huge \sf \fcolorbox{red}{pink}{Solution :)}$

Given ,

Current = 0.1 amp

Resistance = 41.8 ohm

Time = 5 min or 300 sec

We know that ,

$\sf \large \fbox{H = {(I)}^{2} RT}$

Substitute the known values , we get

$\sf \hookrightarrow H = {(0.1)}^{2} \times 41.8 \times 300 \\ \\ \sf \hookrightarrow H = 125.4 \: \: joule$

To convert joule to calorie , divide the joule value by 4.18

$\sf \hookrightarrow H = \frac{125.4}{4.18} \\ \\ \sf \hookrightarrow H = 30 \: \: calorie$

Hence , the heat produced in wire is 30 calorie

______________ Keep Smiling ☺