A student uses the resistance of a known resistor ($1 \Omega$ ) to calibrate a voltmeter and an ammeter using the circuits shown below. The student measures the ratio of the voltage to current to be $1\times {10}^3\Omega$ in circuit
(a) and $0.999\Omega$ in circuit (b). From these measurements, the resistance (in $\Omega$ ) of the voltmeter and ammeter are found to be close to

 When a voltmeter put in series, it still reads potential drop and when an ammeter is connected in parallel, it still shows current through it.
Case a

Let $I=$ current through cell, then potential drop read by voltmeter is $V=I·R_V$ (this is reading of voltmeter) Where, $R_V$ is the resistance of voltmeter In loop$A B$, $V_{AB}=I_1\times 1=I_2\times R_A$ and $I=I_1+I_2$
Where, $R_A$ is the resistance of ammeter We substitute for $I_1$ from above equation to get $\Rightarrow I=I_2{R}_A+I_2=I_2\left(R_A+1\right)$
$\Rightarrow I_2=\frac{I}{\left(R_A+1\right)}$
(this is reading of ammeter)

Now given, $\frac{\text{ voltmeter readin'g }}{\text{ ammeter reading }}=1\times {10}^3=\frac{I{R}_V}{\left(\frac{I}{R_A+1}\right)}$
$\mathrm{So},R_V\left(R_A+1\right)=1000$1000$
Case b

Let $I=$ current through cell, then ammeter reading in this case is I.
Also, in loop A B,
$V_{AB}=I_1\times 1=I_2\times R_V$

As, $I=I_1+I_2=I_2{R}_V+I_2$

$=I_2\left(R_V+1\right)$
So, $I_2=\frac{I}{\left(R_V+1\right)}$
Hence, voltmeter reading is $V=I_2{R}_V$ $=\frac{I{R}_V}{\left(R_V+1\right)}$ (this is reading of voltmeter)
Now given, voltmeter reading $÷$ ammeter reading $=0.999\Omega .$
$\mathrm{So},0.999=\frac{\left[\frac{I{R}_V}{\left(R_V+1\right)}\right]}{I}$

$\Rightarrow 0.999=\frac{R_V}{R_V+1}$

$\mathrm{So},R_V=999\Omega$
$\approx {10}^3\Omega$
Substituting $R_V$ in Eq (i), we get
$R_A=\frac{1}{999}$
or $R_A={10}^{-3}\Omega$