**Lecture Notes :- **

Let us solve Numericals Explaining the range of the readings.

**Q.1. Resistances R _{1} and R_{2} have respectively nominal values of 10Ω and 5 Ω and tolerances of ±5% and ±10%. The range of values for the parallel combination of R_{1 }and R_{2} is**

**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 R_{1 }= 10 **Ω**

Nominal Values (True value ) of R_{2 }= 5 **Ω**

So, the formula for range is ,

Range = TV( True Value) ± Error

Range = True; value; left ( 1pm frac{Error}{TV} right )Range of R_{1} = 10pm 10times frac{5}{100}= 9.5Omega (lower); to ; 10.5 ; Omega (higher)

Range of R_{2} = 5pm 5times frac{10}{100}= 4.5Omega (lower) ; to ; 5.5 ; Omega (higher)** **

The range of values for the parallel combination of R_{1 }and R_{2} is given as,

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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