
If $ 400\Omega $ of resistance is made by adding four $ 100\Omega $ resistance of tolerance $ 5\% $ , then the tolerance of the combination is:
(A) $ 20\% $
(B) $ 5\% $
(C) $ 10\% $
(D) $ 15\% $
Answer
553.2k+ views
Hint: The maximum difference between its actual value and the required value is the tolerance of a resistor and is generally expressed as a percentage plus or minus value. Tolerance is the proportion of error in the resistance of the resistor, or how much more or less you can expect from its stated resistance to be the actual measured resistance of a resistor.
Formula Used The formula to find out the total tolerance is given by
$ T = \dfrac{{\Delta R}}{R} $
Where, $ T $ is the total tolerance
$ \Delta R $ is the tolerance percentage
$ R $ is the resistance of the resistor.
Complete step by step answer:
Let each of the four resistances be $ {R_1} $ , $ {R_2} $ , $ {R_3} $ , and $ {R_4} $
It is also given that the tolerance of each resistor is $ 5\% $
That is
$ \dfrac{{\Delta {R_1}}}{R} = \dfrac{5}{{100}} $
So, we get
$ \Delta {R_1} = 100 \times \dfrac{5}{{100}} = 5\Omega $
It is also provided in the question that
$ {R_1} = {R_2} = {R_3} = {R_4} $
Therefore,
$ \Delta {R_1} = \Delta {R_2} = \Delta {R_3} = \Delta {R_4} = 5\Omega $
Let the net effective resistance be $ {R_e} $
Then,
$ {R_e} = {R_1} + {R_2} + {R_3} + {R_4} $
$ \Rightarrow {R_e} = 100 + 100 + 100 + 100 = 400\Omega $
Now if we apply differential sign on both the sides of the equation, we get
$ \Delta {R_e} = \Delta {R_1} + \Delta {R_2} + \Delta {R_3} + \Delta {R_4} $
Now we will put the value of each term in the given equation
$ \Delta {R_e} = 5 + 5 + 5 + 5 = 20\Omega $
And we have to find out the total tolerance of the combination of resistances
That is, we need to find
$ \dfrac{{\Delta {R_e}}}{{{R_e}}} \times 100 $
Now let us put values of each term
$ \dfrac{{20}}{{400}} \times 100 = 5\% $
Hence the correct option is (B.)
Note:
In an electrical circuit, resistance is a measure of the opposition to current flow. We measure resistance in ohms. Tolerance on the other hand is measured in percentage. It can either be positive or negative.
Formula Used The formula to find out the total tolerance is given by
$ T = \dfrac{{\Delta R}}{R} $
Where, $ T $ is the total tolerance
$ \Delta R $ is the tolerance percentage
$ R $ is the resistance of the resistor.
Complete step by step answer:
Let each of the four resistances be $ {R_1} $ , $ {R_2} $ , $ {R_3} $ , and $ {R_4} $
It is also given that the tolerance of each resistor is $ 5\% $
That is
$ \dfrac{{\Delta {R_1}}}{R} = \dfrac{5}{{100}} $
So, we get
$ \Delta {R_1} = 100 \times \dfrac{5}{{100}} = 5\Omega $
It is also provided in the question that
$ {R_1} = {R_2} = {R_3} = {R_4} $
Therefore,
$ \Delta {R_1} = \Delta {R_2} = \Delta {R_3} = \Delta {R_4} = 5\Omega $
Let the net effective resistance be $ {R_e} $
Then,
$ {R_e} = {R_1} + {R_2} + {R_3} + {R_4} $
$ \Rightarrow {R_e} = 100 + 100 + 100 + 100 = 400\Omega $
Now if we apply differential sign on both the sides of the equation, we get
$ \Delta {R_e} = \Delta {R_1} + \Delta {R_2} + \Delta {R_3} + \Delta {R_4} $
Now we will put the value of each term in the given equation
$ \Delta {R_e} = 5 + 5 + 5 + 5 = 20\Omega $
And we have to find out the total tolerance of the combination of resistances
That is, we need to find
$ \dfrac{{\Delta {R_e}}}{{{R_e}}} \times 100 $
Now let us put values of each term
$ \dfrac{{20}}{{400}} \times 100 = 5\% $
Hence the correct option is (B.)
Note:
In an electrical circuit, resistance is a measure of the opposition to current flow. We measure resistance in ohms. Tolerance on the other hand is measured in percentage. It can either be positive or negative.
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