Question

The resistance of a heater coil is $110\Omega$ . A resistance R is connected parallel with it and combination is joined in series with a resistance of $11\Omega$ to a 220 V main line. The heater operation with a power of 110 W. The value of R in ohm is:
(A) 12.22
(B) 24.42
(C) Negative
(D) That the given value is not correct

Answer 1

In order to solve this question we will solve and find the resistance from the relation of P,V and R. Since voltage is given and resistance is found by us so by the ohm's law current can be calculated. Now there is the resistance of $110\Omega$ parallel to R so the current (I) in $110\Omega$ can be calculated and by subtracting from the total current we can get the current across R now and voltage is already given so the resistance can be found.

Complete step by step solution:
For solving this we need to make the figure of what is the question telling us so we will be making the figure of it as:
According to the question the power given to us is 110W.
So by the relation between P,V and R:
$P = \dfrac{{{V^2}}}{R}$
By putting the values of P and R we will find the value of V:
$110 = \dfrac{{{V^2}}}{{110}}$
On further solving we will get the value of V = 110V.
Now since the values of voltage and resistance are given so we will calculate the value of I.
By ohm's law:
$I = \dfrac{V}{R}$
Putting the values of V and R,
$I = \dfrac{{110}}{{110}}$
On further solving we get:
I = 1A
Now the total current will be:
${i_1}$ = voltage in that circuit/ resistance on that circuit
Since voltage is the same in the whole circuit and resistance is $11\Omega$ .
So according to equation:
${i_1} = \dfrac{{110}}{{11}}$ A
${i_1} = 10A$
The current in the lower circuit will be:
${i_2} = {i_1} - I$
Putting the values of $I$ and ${i_1}$ in this equation:
${i_2} = 10 - 1$
On further solving we get:
${i_2} = 9A$
For finding the R in the lower circuit, as we already know the current and voltage the resistance can be found by the ohm's law;
$V = iR$
Putting the values of V and i
$110 = 9R$
Now by transforming the values we get:
$R = \dfrac{{110}}{9}$
On further solving we get:
$R = 12.22\Omega$
Hence our A option is correct.

Note:
An alternative method for after when the voltage is found:
Where V = 110V
So finding
${I_1} = \dfrac{R}{{110 + R}}I$ …………………….(1)
And for finding
${I_2} = \dfrac{{110}}{{110 + R}}I$ ……………………(2)
Adding equation (1) and (2) we get
$\dfrac{{10R}}{{110 + R}} + \dfrac{{1100}}{{110 + R}} = 10$
Putting the above calculated value of ${I_1}$ which is equal to 1
$1 + \dfrac{{1100}}{{110 + R}} = 10$
On further solving
$1100 = 990R + 9R$
On transferring these to appropriate position we get:
$9R = 110$
Now the final value of R will be equal to 12.22ohm.