Question

An electric stove boils 1kg of water in time 2 min and another stove boils 1kg of water in time 3 min. both electric stoves are designed for the same voltage. when they are joined in parallel, the time required to boil 1 kg of water is​

Answer 1

Answer:

Explanation:

HELLO DEAR,

Given that i) An electric stove boil 1kg water = 2 minute.

ii) And another electric stove boil 1kg of water = 3 minut.

To find the time taken to boil 1kg of water when both i) and ii) electric stove join in parallel to elecric circuit.

Lets get start it.

Let the energy to boil 1kg of water is

H= x.

Then by using formula of Joules Law of Heating H= I^2 R t.

Now for i) elecric stove we have to find resistance of it which we take R1.

So, H= l^2 R1t

x = l^2R1t. ( H= x)

R1= x/ l^2t

where t = 2 min. Given

R1= x/2l^2. ..........iii)

Now for ii) electric stove we have to find the resistance of it which we take R2.

So, H= l^2R2 t

x = l^2 R2 t

R2= x/ l^2 t

where t= 3minut given

R2= x/ 3l^2 .........iv)

Then the electric stove join in parallel means its resistance join in parallel.

Therefore total resistance = > R= (R1 R2)/(R1 + R2)

=> R= x/ 5l^2. ( After solving ,given in pic).

Therefore H= l^2Rt

x= l^2 . x/ 5l^2 t

x = xt/ 5

t = 5 minut. Answer.

I HOPE YOU UNDERSTAND DEAR,

THANKS.

Answer 2

Given that,

Time for first stove = 2 min

Time for second stove = 3 min

Both electric stoves are designed for the same voltage.

Let R₁ and R₂ be the resistance of two stoves and H in the required to boil 1 kg of water if V is the voltage applied to heated

We need to find the value of R₁ and R₂

Using formula of heat

$H=\dfrac{V^2}{R_{1}}t_{1}=\dfrac{V^2}{R_{2}}t_{2}$...(I)

$\dfrac{t_{1}}{R_{1}}=\dfrac{t_{2}}{R_{2}}$

$\dfrac{R_{1}}{R_{2}}=\dfrac{t_{1}}{t_{2}}$.....(II)

When they are joined in parallel,

We need to calculate the time

Using formula of heat

$H=(\dfrac{V^2}{R_{1}}+\dfrac{V^2}{R_{2}})t$

From equation (I)

$\dfrac{V^2}{R_{1}}t_{1}=(\dfrac{V^2}{R_{1}}+\dfrac{V^2}{R_{2}})t$

$\dfrac{t_{1}}{R_{1}}=(\dfrac{1}{R_{1}}+\dfrac{1}{R_{2}})t$

$\dfrac{t_{1}}{t}=\dfrac{R_{1}(R_{2}+R_{1})}{R_{1}R_{2}}$

$\dfrac{t_{1}}{t}=1+\dfrac{R_{1}}{R_{2}}$

$\dfrac{t_{1}}{t}=1+\dfrac{t_{1}}{t_{2}}$

$\dfrac{t_{1}}{t}=\dfrac{t_{2}+t_{1}}{t_{2}}$

$t=\dfrac{t_{2}t_{1}}{t_{2}+t_{1}}$

Put the value into the formula

$t=\dfrac{3\times2}{3+2}$

$t=\dfrac{6}{5}\ min$

Hence, The time required to boil 1 kg of water is $\dfrac{6}{5}$