A galvanometer of resistance 40 Ω gives a deflection of 5 divisions per mA. There are 50 divisions on the scale. The maximum current that can pass through it when a shunt resistance of 2 Ω is connected is

Question

A galvanometer of resistance 40 Ω gives a deflection of 5 divisions per mA. There are 50 divisions on the scale. The maximum current that can pass through it when a shunt resistance of 2 Ω is connected is (A) 210 mA (B) 155 mA (C) 420 mA (D) 75 mA

Tardigrade Admin · Accepted Answer

IG=(50/5)=10 mA; RG=40 Ω, Rs=2 Ω Maximum current, I=(RG+RS/RG)× IG =((40+2)×10/2) =210 mA

Q. $A$ galvanometer of resistance $40 Ω$ gives a deflection of $5$ divisions per $m A$ . There are $50$ divisions on the scale. The maximum current that can pass through it when a shunt resistance of $2 Ω$ is connected is

$210 m A$

$155 m A$

$420 m A$

$75 m A$

$I_{G} = \frac{50}{5} = 10 m A; R_{G} = 40 Ω, R_{s} = 2 Ω$
Maximum current,
$I = \frac{R _{G} + R _{S}}{R _{G}} \times I_{G}$
$= \frac{( 40 + 2 ) \times 10}{2}$
$= 210 m A$

Q. A galvanometer of resistance 40Ω gives a deflection of 5 divisions per mA. There are 50 divisions on the scale. The maximum current that can pass through it when a shunt resistance of 2Ω is connected is

210mA

155mA

420mA

75mA