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Current Electricity

JEE Main level, single correct, medium

366 of 380

Question 366 of 380

A $50\Omega$ resistance is connected to a battery of $5V$ . 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 with in $1%$ of the current without the ammeter in the circuit?

Show answer and solution

Answer: (D) ${r}_{S}=0.5\Omega$ in parallel with the galvanometer

Initially current in the circuit is

I=\dfrac{5}{50}=0.1

and the current in circuit when ammeter is connected,is

1%

of initial current , i.e.,

{I}^{'}=\dfrac{1}{100}\times 0\cdot 1=0.099A

To increase the range of galvanometer, let resistance

{r}_{s}

is connected parallel to it and along with

50\Omega

is connected in series to ammeter. So, equivalent resistance of circuit is:

{R}_{eq}=50+\dfrac{100{r}_{S}}{100+{r}_{S}}

Now, From equation

V={I}^{'}{R}_{eq}

5=0.099\times \left(50+\dfrac{100{r}_{s}}{100+{r}_{s}}\right) \Rightarrow 50+\dfrac{100{r}_{s}}{100+{r}_{s}}=\dfrac{5}{0.099} \Rightarrow 50+\dfrac{100{r}_{s}}{100+{r}_{s}}=50.5050 \Rightarrow \dfrac{100{r}_{s}}{100+{r}_{s}}=50.5050 \Rightarrow {r}_{s}=0.5\Omega

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