An ammeter that has an internal resistance of 100 $Ω$ is used to measure the current through resistor $R_{3}$ as shown in figures below. The percentage error caused due to meter loading is _____%.

0.625

No worries! We‘ve got your back. Try BYJU‘S free classes today!

6.25

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.25

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B 6.25
Looking back into terminals x and y and using Thevanin's equivalent resistance,
$R_{1} = R_{3} + \frac{R_{1} \times R_{2}}{R_{1} + R_{2}} = 1 k + \frac{1 k \times 1 k}{1 k + 1 k} = 1.5 k Ω$

$\frac{I_{m}}{I} = \frac{R_{t}}{R_{t} + R_{m}}$

$∴ \frac{I_{m}}{I} = \frac{1.5 k Ω}{1.5 k Ω + 100 Ω} = \frac{1.5 k}{1.6 k} = 0.9375$

Error due to loading

$= (1 - \frac{I_{m}}{I}) \times 100$

$= (1 - 0.9375) \times 100 = 6.25 %$

Suggest Corrections