Question

A galvanometer of resistance 50 Ω gives full scale deflection with a current of 5 mA
resistance of 0.1 M Ω is connected in series to convert it into volt meter. Find the
range of voltmeter obtained

Answer 1

Answer:

The range of the voltmeter is 0 V to 500.25 V.

Explanation:

Given, resistance of the galvanometer($R_G= 50 \Omega$)

Current passing through the galvanometer($I_G = 5mA = 5\times10^{-3}A$)

Resistance connected in series ($R = 0.1M\Omega = 10^5\Omega$)

To find, voltage of the obtained volt meter($V$).

As we know that, from ohm's law:

Voltage = Current × Resistance.

So, $V = I_G(R_G + R)$ (as the resistances are connected in series).

$V = 5\times10^{-3} A(50\Omega+ 10^5\Omega)$

V = 500.25 V.

Therefore, the range of the voltmeter is 0 V to 500.25 V.

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Answer 2

Answer:

The range of voltmeter is 0v to 500.25v.

Explanation:

Given,

Ig(Current passing through galvanometer) = 5mA.

R(Resistance through galvanometer)  = 50 Ω.

Resistance in series= 10^5 Ω.

We have to find V .

V= Ig(Rg+R)

V=5x10^-3(50+10^5 Ω)

V = 500.25

So the range of voltmeter is 0v to 500.25v

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