JEE Main 2023 — Physics Electromagnetism

The formula to calculate the maximum current through the galvanometer, when a shunt resistance ${r}_{1}$r1​ is connected is given by${({i}_{G})}_{\mathrm{max}}=\frac{{r}_{1}}{{r}_{1}+{R}_{G}}{i}_{\mathrm{max}}...(1)$(iG​)max​=r1​+RG​r1​​imax​...(1)

When a series resistance ${r}_{2}$r2​ is connected, the potential difference across the galvanometer is given by

$$V={({i}_{G})}_{\mathrm{max}}[{R}_{G}+{r}_{2}]...(2)$$V=(iG​)max​[RG​+r2​]...(2)

From equations (1) and (2), it can be written that

$$\begin{matrix}V & = & \frac{{r}_{1}}{{r}_{1}+{R}_{G}}{i}_{\mathrm{max}}({R}_{G}+{r}_{2}) \\ & \Rightarrow & V{r}_{1}+V{R}_{G}={i}_{\mathrm{max}}{r}_{1}{R}_{G}+{i}_{\mathrm{max}}{r}_{1}{r}_{2} \\ & \Rightarrow & (V-{i}_{\mathrm{max}}{r}_{1}){R}_{G}={i}_{\mathrm{max}}{r}_{1}{r}_{2}-V{r}_{1} \\ & \Rightarrow & {R}_{G}=\frac{{i}_{\mathrm{max}}{r}_{1}{r}_{2}-V{r}_{1}}{V-{i}_{\mathrm{max}}{r}_{1}}...(3)\end{matrix}$$V​=⇒⇒⇒​r1​+RG​r1​​imax​(RG​+r2​)Vr1​+VRG​=imax​r1​RG​+imax​r1​r2​(V−imax​r1​)RG​=imax​r1​r2​−Vr1​RG​=V−imax​r1​imax​r1​r2​−Vr1​​...(3)​

Substitute the values of the known parameters into equation (3) to calculate the required galvanometer resistance.

$$\begin{matrix}{R}_{G} & = & \frac{0.250A\times 5\Omega \times 1050\Omega -25V\times 5\Omega }{25V-0.250A\times 5\Omega } \\ & = & 50\Omega \end{matrix}$$RG​​==​25V−0.250A×5Ω0.250A×5Ω×1050Ω−25V×5Ω​50Ω​