Numerical explaning Range of the Readings (05:46)

Lecture Notes :-

Let us solve Numericals Explaining the range of the readings.

Q.1. Resistances R₁ and R₂ have respectively nominal values of 10Ω and 5 Ω and tolerances of ±5% and ±10%. The range of values for the parallel combination of R₁ and R₂ is [ Gate 2001]

a) 3.077 Ω to 3.636 Ω

b) 2.805 Ω to 3.371 Ω

c) 3.237 Ω to 3.678 Ω

d) 3.192 Ω to 3.435 Ω

Ans. Given Nominal values (True value ) of R₁ = 10 Ω

Nominal Values (True value ) of R₂ = 5 Ω

So, the formula for range is ,

Range = TV( True Value) ± Error

Range = True\; value\; \left ( 1\pm \frac{Error}{TV} \right )

Range of R₁ = 10\pm 10\times \frac{5}{100}= 9.5\Omega (lower)\; to \; 10.5 \; \Omega (higher)

Range of R₂ = 5\pm 5\times \frac{10}{100}= 4.5\Omega (lower) \; to \; 5.5 \; \Omega (higher)

The range of values for the parallel combination of R₁ and R₂ is given as,

R_{p}=\frac{R_{1}R_{2}}{R_{1}+R_{2}}

Lower range of R_p = \frac{9.5 \times 4.5 }{9.5 + 4.5} = 3.053 Ω

Upper range of R_p = \frac{10.5 \times 5.5 }{10.5 + 5.5} = 3.609 Ω

The range of R_p = 3.05 Ω to 3.609 Ω

Therefore, option (a) is the correct answer.

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