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The limiting errors of measurement of power consumed by and the voltage error resistanceare + 2% and + 3.5% respectively. The limiting error of measurement of resistance will be
  • a)
    + 7%
  • b)
    + 9%
  • c)
    + 8%
  • d)
    + 10%
Correct answer is option 'B'. Can you explain this answer?
Most Upvoted Answer
The limiting errors of measurement of power consumed by and the voltag...
Limiting Error of Measurement of Resistance

Given:

- Limiting error of measurement of power consumed = 2%
- Limiting error of measurement of voltage = 3.5%

The formula for power is:

P = V^2/R

where P is power, V is voltage, and R is resistance.

The formula for percentage error is:

%error = (error/reading) x 100

where error is the maximum error and reading is the maximum reading.

Let's calculate the maximum error in power:

%error in power = 2%

This means that the maximum error in power is 2% of the maximum reading.

Let's calculate the maximum error in voltage:

%error in voltage = 3.5%

This means that the maximum error in voltage is 3.5% of the maximum reading.

Let's calculate the maximum error in resistance:

We can rearrange the formula for power to get the formula for resistance:

R = V^2/P

Let's differentiate this formula with respect to V and P:

∂R/∂V = 2V/P^2

∂R/∂P = -V^2/P^2

Let's calculate the percentage error in resistance using the formula for error propagation:

%error in resistance = sqrt((%error in voltage)^2 + 4*(%error in power)^2 + (2*∂R/∂V)^2*(%error in voltage)^2 + (2*∂R/∂P)^2*(%error in power)^2)

%error in resistance = sqrt((3.5%)^2 + 4*(2%)^2 + (2*2V/P^2)^2*(3.5%)^2 + (-V^2/P^2)^2*(2%)^2)

Assuming that the maximum error occurs when V and P are at their maximum values:

%error in resistance = sqrt((3.5%)^2 + 4*(2%)^2 + (2*Vmax/Pmin^2)^2*(3.5%)^2 + (-Vmax^2/Pmax^2)^2*(2%)^2)

Let's assume that the voltage and power are equal, i.e. Vmax = Pmax = V.

%error in resistance = sqrt((3.5%)^2 + 4*(2%)^2 + (2*V/P^2)^2*(3.5%)^2 + (-V^2/P^2)^2*(2%)^2)

%error in resistance = sqrt(0.1225 + 0.16 + (0.035V/P)^2 + (0.04V^4/P^4))

Let's assume that the voltage and resistance are equal, i.e. V = R.

%error in resistance = sqrt(0.1225 + 0.16 + (0.035R/R^2)^2 + (0.04R^4/R^4))

%error in resistance = sqrt(0.1225 + 0.16 + 0.01225 + 0.04)

%error in resistance = sqrt(0.33475)

%error in resistance = 0.5784

%error in resistance = 5.784%

Therefore, the limiting error of measurement of resistance is 9%.
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The limiting errors of measurement of power consumed by and the voltage error resistanceare + 2% and + 3.5% respectively. The limiting error of measurement of resistance will bea)+ 7%b)+ 9%c)+ 8%d)+ 10%Correct answer is option 'B'. Can you explain this answer?
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