Question

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

• A. $10^2$ and $10^{-2}$
• B. $10^3$ and $10^{-3}$
• C. $10^{-2}$ and $10^2$
• D. $10^{-2}$ and $10^3$

Answer 1

Answer: (B)

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

$V=I\cdot Rv$ (this is reading of voltmeter)

Where, $R_v$ is the resistance of voltmeter

In loop $AB$,

$V_{AB}=I_1\times 1=I_2\times R_A$ and $I=I_1+I_2$

Where, $R_A$ is the resistance of ammeter We substitute for $I_1$ from above equation to get

$\Rightarrow I=I_2{R}_A+I_2=I_2(R_A+1)$

$\Rightarrow I_2=\frac{I}{(R_A+1)}$

(this is reading of ammeter)

Now given,

$\frac{\text{voltmeter reading}}{\text{ammeter reading}}=1\times 10^3$

$=\frac{I{R}_v}{(\frac{I}{R_A+1})}$

So, $R_V(R_A+1)=1000...(i)$

Case b



Let $I=$ current through cell, then ammeter reading in this case is $I$.

Also, in loop $AB$,

$V_{AB}=I_1\times 1=I_2\times R_V$

As $I=I_1+I_2=I_2{R}_v+I_2$

So, $I_2=\frac{I}{(R_v+1})$

Hence, voltmeter reading is $V=I_2{R}_V$

$=\frac{I{R}_V}{(R_V+1)}$ (this is reading of voltmeter)

Now given, voltmeter reading $÷$ ammeter reading $=0.999\Omega$

So, $0.999=\frac{\left[\frac{I{R}_V}{\left(R_V+1\right)}\right]}{I}$

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

So, $R_V=999\Omega ...(ii)$

$\approx 10^3\Omega$

Substituting $R_v$ in Eq $(i)$, we get

$R_A=\frac{1}{999}$

or $R_A=10^{-3}\Omega$