Question

d. High potential
9. A potential difference of 20V is applied across a coil. The amount
of heat generated in it is 800cal/s. Value of resistance will be
12 0​

Answer 1

${ \underline{ \sf{AnswEr :}}}$

Here, we're given with the values of potential difference applies accross the coil and amt. of heat generated accross the coil.

That is,

• Potential difference (V) = 20 V
• Heat generated (H) = 800 cal/s

${ \underline{ \sf{Step-by-step \: explanation :}}}$

We need to apply the concept of Electric Power over here !! In easy word's we can define it as the rate of energy consumption in an electrical circuit. Also, SI unit of Power is watt.

Also, as we've given the value of heat generated,that is, 800 cal/s. We need to convert it into J/s, respectively.

• H = 800 cal/s = 800 × 4.2 (1 cal = 4.2J)

Power [in terms of PV and Heat] is given by the formula,

$: \implies \sf \: \: P = \dfrac{{ V}^{2}}{R}$

Simplifying the values,

$: \implies \sf \: \: R= \dfrac{{ V}^{2}}{P}$

Substituting the values,

$: \implies \sf \: \: R= \dfrac{{(20)}^{2}}{800 \times 4.2}$

$: \implies \sf \: \: R= \dfrac{400}{800 \times 4.2}$

$: \implies \sf \: \: R= \dfrac{1}{2 \times 4.2}$

$: \implies \sf \: \: R= \dfrac{1}{8.4}$

$: \implies \sf \: \: { \underline{ \boxed{ \purple{ \frak{ R= 0.12 \: ohm}}}}} \: \bigstar$

Hence, the value of resistance is 0.12 ohm.

Answer 2

Answer:

$\huge{ \sf{ \boxed{ \underline{ \mathfrak{ \star{ \purple{answer}{} }{} }{} }{} }{} }{} }{}$

Given:-

From the question it is given that A potential difference is applied on through a coil. And Heat is liberated

This Means That the,

★Potential Difference (V) = 20V

★Heat generated = 800cal/s

Here we apply the theoretical concept of Electric power.

Here we are given with 800cal/s We need to convert to J/s

• H = 800 cal/s = 800*4.2 (Since 1cal/s = 4.2J/s)

Now we have converted it Lets do the further,

• Power = V2/R

• By simplifying it

• R =V2/P

$\sf{ \bold{ \green{substitute \: the \: values}{} }{} }{}$

$R = \frac{ {20}^{2} }{800 \times 4.2 } \\ \\ R = \frac{ {400}^{} }{800 \times 4 } \\ \\ R = \frac{1}{2 \times 8.4} \\ \\ R = \frac{1}{8.4} = 0.12 \\ \\ \sf{ \red{value \: of \: Resistence \: = 0.12 \: ohm}{} }{}$

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