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. If the full scale deflection current of the galvanometer is $\frac{x}{100}A$. Find $x$.

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= $\frac{5}{4}0.04=0.05A$

therefore full scale reading of galvanometer is 0.05A . Therefore $x=5$