Current Electricity

JEE Main level, numerical answer, medium

336 of 380

Question 336 of 380

When a resistance of 5Ω5\Omega is shunted with a moving coil galvanometer, it shows a full scale deflection for a current of 250mA,250mA, however when 1050Ω1050\Omega resistance is connected with it in series, it gives full scale deflection for 2525 volt. The resistance of galvanometer is_________\Omega .

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Answer: 50

The formula to calculate the maximum current through the galvanometer, when a shunt resistance r1{r}_{1} is connected is given by(iG)max⁡=r1r1+RGimax⁡...(1){\left({i}_{G}\right)}_{\max }=\dfrac{{r}_{1}}{{r}_{1}+{R}_{G}}{i}_{\max }...\left(1\right)
When a series resistance r2{r}_{2} is connected, the potential difference across the galvanometer is given by
V=(iG)max⁡[RG+r2]...(2)V={\left({i}_{G}\right)}_{\max }\left[{R}_{G}+{r}_{2}\right]...\left(2\right)
From equations (1) and (2), it can be written that
V=r1r1+RGimax⁡(RG+r2)⇒Vr1+VRG=imax⁡r1RG+imax⁡r1r2⇒(V−imax⁡r1)RG=imax⁡r1r2−Vr1⇒RG=imax⁡r1r2−Vr1V−imax⁡r1...(3)\begin{matrix}V & = & \dfrac{{r}_{1}}{{r}_{1}+{R}_{G}}{i}_{\max }\left({R}_{G}+{r}_{2}\right) \\ & \Rightarrow & V{r}_{1}+V{R}_{G}={i}_{\max }{r}_{1}{R}_{G}+{i}_{\max }{r}_{1}{r}_{2} \\ & \Rightarrow & \left(V-{i}_{\max }{r}_{1}\right){R}_{G}={i}_{\max }{r}_{1}{r}_{2}-V{r}_{1} \\ & \Rightarrow & {R}_{G}=\dfrac{{i}_{\max }{r}_{1}{r}_{2}-V{r}_{1}}{V-{i}_{\max }{r}_{1}}...\left(3\right)\end{matrix}
Substitute the values of the known parameters into equation (3) to calculate the required galvanometer resistance.
RG=0.250A×5Ω×1050Ω−25V×5Ω25V−0.250A×5Ω=50Ω\begin{matrix}{R}_{G} & = & \dfrac{0.250A\times 5\Omega \times 1050\Omega -25V\times 5\Omega }{25V-0.250A\times 5\Omega } \\ & = & 50\Omega \end{matrix}

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