A galvanometer of resistance 40 Omega gives a | vedclass

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Question

(A) Given: Galvanometer resistance $R_G = 40 \Omega$, Shunt resistance $R_S = 2 \Omega$, Sensitivity $= 10 	ext{ div/mA}$, Total divisions $n = 50$.T

Vedclass · Accepted Answer

$105$

Vedclass · Answer

(A) Given: Galvanometer resistance $R_G = 40 \Omega$, Shunt resistance $R_S = 2 \Omega$, Sensitivity $= 10 	ext{ div/mA}$, Total divisions $n = 50$.The maximum current that the galvanometer can measure without shunt is $I_G = \frac{50 	ext{ div}}{10 	ext{ div/mA}} = 5 	ext{ mA}$.When a shunt $R_S$ is connected in parallel, the total current $I$ is divided between the galvanometer and the shunt.Using the principle of parallel circuits, the voltage across the galvanometer and shunt is equal: $I_G R_G = I_S R_S$.Since $I = I_G + I_S$, we have $I_S = I - I_G$.Substituting this: $I_G R_G = (I - I_G) R_S$.Rearranging for $I$: $I = I_G \left( 1 + \frac{R_G}{R_S} ight) = I_G \left( \frac{R_G + R_S}{R_S} ight)$.Substituting the values: $I = 5 	ext{ mA} 	imes \left( \frac{40 + 2}{2} ight) = 5 	imes \frac{42}{2} = 5 	imes 21 = 105 	ext{ mA}$.

$A$ galvanometer of resistance $40 \Omega$ gives a deflection of $10 \text{ divisions per } mA$. There are $50 \text{ divisions}$ on the scale. The maximum current that can pass through the circuit when a shunt resistance of $2 \Omega$ is connected is: (in $\text{ mA}$)

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