Question
Question: If \( 400\Omega \) of resistance is made by adding four \( 100\Omega \) resistance of tolerance \( 5...
If 400Ω of resistance is made by adding four 100Ω resistance of tolerance 5% , then the tolerance of the combination is:
(A) 20%
(B) 5%
(C) 10%
(D) 15%
Solution
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=RΔR
Where, T is the total tolerance
ΔR is the tolerance percentage
R is the resistance of the resistor.
Complete step by step answer:
Let each of the four resistances be R1 , R2 , R3 , and R4
It is also given that the tolerance of each resistor is 5%
That is
RΔR1=1005
So, we get
ΔR1=100×1005=5Ω
It is also provided in the question that
R1=R2=R3=R4
Therefore,
ΔR1=ΔR2=ΔR3=ΔR4=5Ω
Let the net effective resistance be Re
Then,
Re=R1+R2+R3+R4
⇒Re=100+100+100+100=400Ω
Now if we apply differential sign on both the sides of the equation, we get
ΔRe=ΔR1+ΔR2+ΔR3+ΔR4
Now we will put the value of each term in the given equation
ΔRe=5+5+5+5=20Ω
And we have to find out the total tolerance of the combination of resistances
That is, we need to find
ReΔRe×100
Now let us put values of each term
40020×100=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.