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  • A galvanometer having 50 divisions provided with a variable shunt S is used to measure the current when connected in series with a resistance of <img alt="90\Omega" src="/latex-image/?90%5COmega" /

A galvanometer having 50 divisions provided with a variable shunt S is used to measure the current when connected in series with a resistance of 90\Omega and a battery of internal resistance 10\Omega. It is observed that when the shunt resistances are10\Omega and 50\Omega respectively, the deflection are respectively 9 and 30 divisions. The resistance of the galvanometer is:

Option: 1

132\Omega


Option: 2

233\Omega


Option: 3

204\Omega


Option: 4

250\Omega


Answers (1)

best_answer

I=\frac{\varepsilon}{\left(90+10+\frac{S G}{S+G}\right)}=\frac{\varepsilon}{\left(100+\frac{S G}{S+G}\right)} ______________(i)

applying Kirchhoff’s law

We get, i_g=\frac{I S}{S+G}

Let i_{g}= I_{1} for S = 10\Omega andig=I_{2} for s=50\Omega

\frac{\mathrm{I}_1}{\mathrm{I}_2}=\frac{\left(\frac{10}{10+G}\right) \times\left(\frac{\varepsilon}{100+\frac{10 G}{10+G}}\right)}{\left(\frac{50}{50+G}\right) \times\left(\frac{\varepsilon}{100+\frac{50 G}{50+G}}\right)} \Rightarrow \frac{11}{12}=\frac{100+3 G}{100+11 G}

\therefore deflection is proportional to the current

\Rightarrow \quad \frac{9}{30}=\frac{100+3 \mathrm{G}}{100+11 \mathrm{G}}

solving we get

\mathrm{G}=233.3 \Omega

Posted by

Anam Khan

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