Galvoltmeter!

Electricity and Magnetism Level 2

A galvanometer coil has a resistance 90 ohm and full scale deflection current 10 mA. A 910 ohm resistance is connected in series with the galvanometer to make a voltmeter. If the least count of the voltmeter is 0.1V, what is the number of divisions on the scale?


The answer is 100.

Rohan Bansal by rohan bansal , 22, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Lu Chee Ket
Dec 14, 2015

V = I R = 10 m A × \times 90 Ω \Omega = 0.9 V {Full deflection}

90 Ω 910 Ω + 90 Ω = 0.9 V 10 V \dfrac{90 \Omega}{910 \Omega + 90 \Omega} = \dfrac{0.9 V}{\boxed{10 V}}

0.9 V for full deflection becomes 10 V for full deflection with 910 Ω 910 \Omega connection in series with it.

Number of divisions = 10 V 0.1 V \dfrac{10 V}{0.1 V} = 100.

Answer: 100 \boxed{100}

3 Helpful 0 Interesting 0 Brilliant 1 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...