Question

The observations of voltage and current in an electric circuit are V=(20±0.1) volt
and I=(4±0.25) Ampere respectively. Find the percentage error in the measurement
of resistance in the circuit.

Answer 1

The observations of voltage and current in an electric circuit are V = (20±0.1) volt and I = (4±0.25) Ampere respectively. The percentage error in the measurement of resistance in the circuit is 6.75%.

• Given,
• V = (20±0.1) volt
• I = (4±0.25) Ampere
• V = IR
• R = V/I
• ln R = ln V ± ln I
• dR/R = dV/V ± dI/I
• dR/R = 0.1/20 ± 0.25/4
• dR/R = 0.005 ± 0.0625
• dR/R = 0.0675
• dR/R = 0.0675 * 100
• =6.75%

Answer 2

Answer:

The percentage error of resistance in circuit is 0.35 %

Explanation:

Given as :

From the observation

voltage = v = $20\pm 0.1$  volt

Current = I = $4\pm 0.25$  Amp

Let The % error for resistance measurement = R%

According to question

∵  Resistance = $\dfrac{voltage}{current}$

% R = $\dfrac{\Delta R}{R}$ × 100

Or, $\dfrac{\Delta R}{R}$ × 100 = $\dfrac{\Delta V}{V}$ × 100 + $\dfrac{\Delta I}{I}$ × 100

Or, $\dfrac{\Delta R}{R}$ × 100 = 0.1 + 0.25

Or, $\dfrac{\Delta R}{R}$ × 100 = 0.35

or, $\dfrac{\Delta R}{R}$ = 0.35 %

Hence, The percentage error of resistance in circuit is 0.35 % Answer