Answer
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Hint: Firstly, we will use the formula for power. Then using the joule’s law of heating, we will write $P = VI$ . Then we will use the ohm’s law $V = IR$ . Then we will put the value of I in the equation of power. We will get to know the value of power in terms of potential difference and current. After calculating the resistance of the hot lamp, we will solve the resistance of the cold lamp by dividing 10.
Complete step by step solution
The amount of work done to carry 1 coulomb charge through a potential difference of 1 volt is called 1 joule.
Again, we can define power as the rate of doing work in unit time.
$\therefore Power(P) = \dfrac{W}{t} = \dfrac{{VIt}}{t}$ [ $\because W = $ electrical work done in the circuit, I= electrical source sends current to the circuit, t= time]
$ \Rightarrow P = VI....(i)$
If V be the terminal potential difference and I be the current flowing through an external resistance R, then according to Ohm’s law, $V = IR....(ii)$
Using Joule’s law of heating,
Putting the value of equation (ii) in (i) we get $P = {I^2}R.....(iii)$
Now $ \Rightarrow P = (V \times \dfrac{V}{R})$
$P = \dfrac{{{V^2}}}{R}....(iv)$
Given that, the power of the lamp is (P) = 100W.
potential difference (V) is =200V
let the resistance be R
now putting those given values in equation (iv) we get
${R_{hot}} = \dfrac{{{{200}^2}}}{{100}} = 400\Omega $ when operating
It is given that the resistance of a hot lamp is 10 times the temperature of a cold lamp.
Let the resistance of cold lamp is ${R_{cold}}$
Hence ${R_{cold}} = \dfrac{{400}}{{10}} = 40\Omega $
The required solution is $40\Omega $ (option-A)
Note One may think that the option is C. But this is incorrect because we have to calculate the resistance when the coil isn’t being used, i.e. when it is cold. So, we have to divide the hot resistance with 10 to get the cold resistance.
Complete step by step solution
The amount of work done to carry 1 coulomb charge through a potential difference of 1 volt is called 1 joule.
Again, we can define power as the rate of doing work in unit time.
$\therefore Power(P) = \dfrac{W}{t} = \dfrac{{VIt}}{t}$ [ $\because W = $ electrical work done in the circuit, I= electrical source sends current to the circuit, t= time]
$ \Rightarrow P = VI....(i)$
If V be the terminal potential difference and I be the current flowing through an external resistance R, then according to Ohm’s law, $V = IR....(ii)$
Using Joule’s law of heating,
Putting the value of equation (ii) in (i) we get $P = {I^2}R.....(iii)$
Now $ \Rightarrow P = (V \times \dfrac{V}{R})$
$P = \dfrac{{{V^2}}}{R}....(iv)$
Given that, the power of the lamp is (P) = 100W.
potential difference (V) is =200V
let the resistance be R
now putting those given values in equation (iv) we get
${R_{hot}} = \dfrac{{{{200}^2}}}{{100}} = 400\Omega $ when operating
It is given that the resistance of a hot lamp is 10 times the temperature of a cold lamp.
Let the resistance of cold lamp is ${R_{cold}}$
Hence ${R_{cold}} = \dfrac{{400}}{{10}} = 40\Omega $
The required solution is $40\Omega $ (option-A)
Note One may think that the option is C. But this is incorrect because we have to calculate the resistance when the coil isn’t being used, i.e. when it is cold. So, we have to divide the hot resistance with 10 to get the cold resistance.
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