JEE MAIN - Physics (2016 - 9th April Morning Slot - No. 1)
A 50 $$\Omega $$ resistance is connected to a battery of 5 V. A galvanometer of resistance 100 $$\Omega $$ is to be used as an ammeter to measure current through the resistance, for this a resistance rs is connected to the galvanometer. Which of the following connections should be employed if the measured current is within 1% of thecurrent without the ammeter in the circuit ?
rs = 0.5 $$\Omega $$ in parallel with the galvanometer
rs = 0.5 $$\Omega $$ in series with the galvanometer
rs = 1 $$\Omega $$ in series with galvanometer
rs =1 $$\Omega $$ in parallel with galvanometer
Explanation
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Current in the circuit without ammeter
I $$=$$ $${5 \over {50}} = 0.1$$ A
$$ \therefore $$ With ammeter current $$=$$ 0.1 $$ \times $$ $${{99} \over {100}}$$ = 0.099 A
With ammeter equivalent resistance,
Req = 50 + $${{100\,{r_s}} \over {100 + {r_s}}}$$
$$ \therefore $$ 0.099 = $${5 \over {50 + {{100\,{r_s}} \over {100 + {r_s}}}}}$$
$$ \Rightarrow $$ 50 + $${{100\,{r_s}} \over {100 + {r_s}}}$$ = $${5 \over {0.099}}$$
$$ \Rightarrow $$ $${{100\,{r_s}} \over {100 + {r_s}}} = 0.5$$
$$ \Rightarrow $$ 100 rs = 50 + 0.5rs
$$ \Rightarrow $$ 99.5 rs = 50
$$ \Rightarrow $$ rs $$=$$ $${{50} \over {99.5}} = 0.5\,\Omega $$
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