$A$ moving coil galvanometer has a resistance of $50\,\Omega$ and gives full-scale deflection for $10\,mA$. How could it be converted into an ammeter with a full-scale deflection for $1\,A$?

  • A
    $50/99\,\Omega$ in series
  • B
    $50/99\,\Omega$ in parallel
  • C
    $0.01\,\Omega$ in series
  • D
    $0.01\,\Omega$ in parallel

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To find the resistance of a galvanometer by the half-deflection method,the experimental data obtained are given in the table below:
$S. No.$Resistance $R \ (\Omega)$Deflection $(\theta)$Shunt $S \ (\Omega)$Half deflection $(\theta / 2)$Galvanometer resistance $(G)$
$1$$3300$$30$$80$$15$$G_1$
$2$$5000$$20$$80$$10$$G_2$

From the above data,the galvanometer resistance $G$ will be near to: (in $\Omega$)

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The current flowing through a coil of resistance $900 \, \Omega$ is to be reduced by $90 \, \%$. What value of shunt should be connected across the coil in $\Omega$?

An ammeter with a resistance of $0.018 \, \Omega$ can measure $1 \, A$. What shunt resistance is required to enable it to measure $10 \, A$?

Given below are two statements: One is labelled as Assertion $A$ and the other is labelled as Reason $R$.
Assertion $A:$ For measuring the potential difference across a resistance of $600\,\Omega$,the voltmeter with resistance $1000\,\Omega$ will be preferred over a voltmeter with resistance $4000\,\Omega$.
Reason $R:$ $A$ voltmeter with higher resistance will draw smaller current than a voltmeter with lower resistance.
In the light of the above statements,choose the most appropriate answer from the options given below.

An ammeter with a resistance of $1\, \Omega$ can measure up to $10\, mA$. To convert it into a voltmeter that can measure up to $10\, V$,what resistance (in $\Omega$) must be connected in series?

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