Question

The actual value of resistance R shown in the figure is $20\mathrm{\Omega }$. This is measured is an experiment as shown using the standard formula $R={\displaystyle \frac{V}{I}}$, where V and I are the readings of the voltmeter and ammeter respectively. If the measured value of R is $10\mathrm{\%}$ less, then the internal resistance of the voltmeter is (in ohms)? (Neglect resistance of ammeter).

Answer 1

Hint: First determine the measured effective resistance value of the circuit given that it is $10\mathrm{\%}$ less than the actual value. Then, obtain an expression for the effective resistance of the circuit which involves the resistor and the voltmeter being parallel to each other. Equate the two expressions and arithmetically solve for the internal resistance of the voltmeter.

Formula Used:
Effective resistance of two resistors in parallel: $R_{eff}={\displaystyle \frac{R_1{R}_2}{R_1+R_2}}$

Complete Step-by-Step Solution:
We are given that the actual value of resistance R is $20\mathrm{\Omega }$.
But the measured value is $10\mathrm{\%}$ less than the actual value, i.e.,
$R_{measured}=R-{\displaystyle \frac{10}{100}}R=R-0.1R=0.9R$
$\Rightarrow R_{measured}=0.9\times 20=18\mathrm{\Omega }$.

We are given that this reduction in the measured value is due to the internal resistance $R_V$ of the voltmeter (we neglect resistance of ammeter). The voltmeter V and the resistance R are connected in parallel to each other. We can solve for the internal resistance of the voltmeter by obtaining an expression for the effective resistance of the voltmeter and the resistor, i.e.,
$R_{eff}={\displaystyle \frac{R\times R_V}{R+R_V}}$.

But we know that the effective resistance of the setup is the measured value of the resistance, i.e.,
$\Rightarrow R_{measured}={\displaystyle \frac{R\times R_V}{R+R_V}}$
$\Rightarrow 18={\displaystyle \frac{20R_V}{20+R_V}}$
$\Rightarrow 18(20+R_V)=20R_V$
$\Rightarrow 20R_V–18R_V=360$
$R_V={\displaystyle \frac{360}{2}}=180\mathrm{\Omega }$

Therefore, the internal resistance of the voltmeter is found to be $180ohms$.

Note:
Remember that a series circuit is a voltage divider whereas a parallel circuit is a current divider. We add up all the resistances in a series circuit because the resistances are arranged in such a way that the current has only one path to take. However, when resistances are connected in parallel, there are multiple paths for the current to pass through but have the same potential difference across them. This results in a net resistance that is lower than any of the individual resistances.

This is also the same reason why the voltmeter might read a lower voltage if a resistance is connected in parallel to it. This is because the internal resistance of the voltmeter is generally very high and if a relatively low resistance is introduced in a parallel branch, most of the current gets diverted to the path of least resistance and the voltmeter receives relatively lesser current and is thus able to process it and show a lesser voltage.