Question

A galvanometer (coil resistance $99\Omega$) is converted into an ammeter using a shunt of $1\Omega$ and connected as shown in figure $1$. The ammeter reads $3\text{A}$. 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 $2$. Its reading is found to be ${\left(\frac{4}{5}\right)}^{th}$ of the full scale reading.

• A. Internal resistance of the cell $r=1.01\Omega$.
  
• B. Range of the ammeter is $5\text{A}$.
• C. Range of the voltmeter is $16\text{V}$.
  
• D. Full scale deflection current of the galvanometer is $0.05\text{A}$
  

Answer 1

Answer: (A), (B), (D)

The correct options are
A Internal resistance of the cell $r=1.01\Omega$.

B Range of the ammeter is $5\text{A}$.
D Full scale deflection current of the galvanometer is $0.05\text{A}$


For ammeter,
$99I_g=(I-I_g)1I=100I_g.....\left(1\right)$
$I_g$ is the full scale deflection current of the galvanometer and $I$ is the range of ammeter.
For the circuit in figure $1$,
$\frac{12}{2+r+\frac{99\times 1}{99+1}}=3\text{A}\Rightarrow r=1.01\Omega$

For voltmeter, range
$V=I_g(99+101)$
$V=200I_g.....\left(2\right)$
Also, resistance of the voltmeter
$=99+101=200\Omega$
In figure $2$, resistance across the terminals of the battery-
$R_1=r+\frac{200\times 2}{202}=2.99\Omega$
$\therefore$ Current drawn from the battery is
$I_1=\frac{12}{2.99}=4.01\text{A}$
$\therefore$ Voltmeter reading
$\frac{4}{5}V=12-I_{1}r=12-4.01\left(1.01\right)$
$V=7.96\times \frac{5}{4}=9.95\text{Volt}$
Using (2), $I_g=\frac{9.95}{200}=0.05\text{A}$
Using (1), range of the ammeter will be
$I=100I_g$
$\therefore$ Range of ammeter $=100\times 0.05=5\text{A}$