Question

A galvanometer (coil resistance $99\Omega$) is converted into a ammeter using a shunt of $1\Omega$ and connected as shown in the figure $\left(i\right)$. The ammeter reads $3A$. The same galvanometer is converted into a voltmeter by connecting a resistance of $101\Omega$ in series. This voltmeter is connected as shown in figure $\left(ii\right)$. Its reading is found to be $4/5$ of the full scale reading. Find the range of the ammeter and voltmeter

Answer 1

when connected as ammeter effective resistance of circuit$R=r+99//1+2=r+0.99+2$
current in circuit = $\frac{12}{R}=3\Rightarrow \frac{12}{3}=R\Rightarrow 4=r+0.99+2\Rightarrow r=1.01\Omega$
consider the voltmeter circuit,
current in circuit = $\frac{12}{r+2}=4A$
voltage across 2 ohm resistance = $\frac{2}{2+1}12=8V$
this is $4/5$th of full scale reading.
Therefore full scale reading of voltmeter= $\frac{5}{4}8=10V$
current passing through the voltmeter= $\frac{2}{200+2}4\approx 0.04A$
this is $4/5$th of maximum current allowed in the galvanometer when used as voltmeter.
therefore ,
maximum current allowed in galvanometer when used as voltmeter= $\frac{5}{4}0.04=0.05A$

when connected as ammeter, maximum current allowed in ammeter is $0.05A$
that is a maximum of 0.05A is allowed through the 99 ohm branch of ammeter.
Therefore total maximum current allowed through ammeter is = $I_{max}\frac{1}{1+99}=0.05A\Rightarrow I_{max}=5A$
therefore, full scale ammeter reading is 5A