Question 404: Current Electricity

Question

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 r
s
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 ?

Accepted Answer

Current in the circuit without ammeter
I $=$ ${5 \over {50}} = 0.1$ A
$ 	herefore $   With ammeter current $=$ 0.1 $ 	imes $ ${{99} \over {100}}$ = 0.099 A
With ammeter equivalent resistance,
R
eq
= 50 + ${{100\,{r_s}} \over {100 + {r_s}}}$
$ 	herefore $   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 r
s
= 50 + 0.5r
s
$ \Rightarrow $   99.5 r
s
= 50
$ \Rightarrow $   r
s
$=$ ${{50} \over {99.5}} = 0.5\,\Omega $

Answer

r
s
= 0.5 $\Omega $ in parallel with the galvanometer

Answer

r
s
= 0.5 $\Omega $ in series with the galvanometer

Answer

r
s
= 1 $\Omega $ in series with galvanometer

Answer

r
s
=1 $\Omega $ in parallel with galvanometer

Current Electricity

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