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A galvanometer having 50 divisions provided with a variable shunt s which 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 are 10 \Omega and 50 \Omega,the deflection are respectively 9 and 30 division. The resistance of the Galvanometer is 

Option: 1

233.3 \Omega


Option: 2

263.3 \Omega


Option: 3

230 \Omega


Option: 4

260 \Omega


Answers (1)

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Current in the main circuit is given as
\mathrm{I=\frac{E}{90+10+\frac{S G}{S+G}}=\frac{E}{100+\frac{S G}{S+G}}}.

Applying Kirchosf's law:
\mathrm{we\, get\,\: i_{g}=\frac{I \cdot S}{S+G} }
\mathrm{\Rightarrow i_{g}=\frac{S}{S+G} \cdot \frac{E}{100+\frac{S G}{S+G}} }

Let \mathrm{i_{g}=i_{1}} for \mathrm{S=10 \Omega} and \mathrm{i_{g}=i_{2}} for \mathrm{S= 50 \Omega}
\mathrm{\frac{i_{1}}{i_{2}}=\frac{\left(\frac{10}{10+G}\right) \cdot \frac{E}{100+\frac{10 G}{10+G}}}{\left(\frac{50}{50+G}\right) \times \frac{E}{100+\frac{50 G}{50+G}}} \Rightarrow \frac{i_{1}}{i_{2}}=\frac{100+3 G}{100+11 G}}.

Since deflection is propotional to the current \mathrm{\Rightarrow \frac{9}{30}=\frac{100+3G}{100+11G} }.
\mathrm{\Rightarrow G =233.3 \Omega }.

 

Posted by

Ritika Harsh

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