An ammeter having resistance R₀ is connected to measure the current intensity in a circuit with a resistance R. What relative error will be made if connection of the ammeter does not change the current intensity in the circuit? The voltage across the ends of the circuit is kept constant.

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

$\frac{R}{R_{0}}$

$\frac{R_{0}}{R}$

$\frac{1}{1 + \frac{R}{R_{0}}}$

$\frac{1}{1 + \frac{R_{0}}{R}}$

answer is C.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

Before the ammeter is connected. The current through the circuit is

$I_{1} = \frac{V}{R}$

When an ammeter of resistance $R_{0}$ is connected in series to the circuit the current will become

$I_{2} = \frac{V}{R + R_{0}}$

The fractional error is $e = \frac{I_{1} - I_{2}}{I_{1}}$

$e = 1 - \frac{I_{2}}{I_{1}} = 1 - \frac{R}{R_{0} + R} = \frac{R_{0}}{R_{0} + R} = \frac{1}{1 + R / R_{0}} .$

When $R_{0} << R$ the error may be neglected

JEE

NEET

Foundation JEE

Foundation NEET

CBSE