Question 330: 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 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 ?

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

rs = 0.5 $$\Omega $$ in parallel with the galvanometer

Answer

rs = 0.5 $$\Omega $$ in series with the galvanometer

Answer

rs = 1 $$\Omega $$ in series with galvanometer

Answer

rs =1 $$\Omega $$ in parallel with galvanometer

Current Electricity

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