A galvanometer of resistance 100 when connected in series with 400 measures a voltage of upto 10 ~V . The value of resistance required to convert the galvanometer into ammeter to read upto 10 ~A is x × 10^ -2 . The

Correct answer: C. 20 i_ g =(10)/(400+100)=20 × 10^ -3 ~A For ammeter Let shunt resistance =S aligned & i_ g R=(i-i_ g ) S & 20 × 10^ -3 × 100=10 ~S & ~S=20 × 10^ -2 aligned

JEE Main 20245 AprilEvening Shift

Exam: JEE Main
Year: 2024
Date: 5 April
Shift: Evening Shift
Subject: Physics
Chapter: Current Electricity

JEE Main 2024 · 5 April · Evening Shift · Physics · Current Electricity

A galvanometer of resistance 100 when connected in series with 400...

A galvanometer of resistance

100 \Omega

when connected in series with

400 \Omega

measures a voltage of upto

10 \mathrm{~V}

. The value of resistance required to convert the galvanometer into ammeter to read upto

10 \mathrm{~A}

x \times 10^{-2} \Omega

. The value of

x

is :

Options

A $2$
B $800$
C $20$
D $200$

Correct answer

C. $20$

Step-by-step solution

\mathrm{i}_{\mathrm{g}}=\frac{10}{400+100}=20 \times 10^{-3} \mathrm{~A}

For ammeter Let shunt resistance

=\mathrm{S}

\begin{aligned} & \mathrm{i}_{\mathrm{g}} \mathrm{R}=\left(\mathrm{i}-\mathrm{i}_{\mathrm{g}}\right) \mathrm{S} \\ & 20 \times 10^{-3} \times 100=10 \mathrm{~S} \\ & \mathrm{~S}=20 \times 10^{-2} \Omega \end{aligned}

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