Question

A voltmeter rated $150V,20k\Omega $ when connected in series with a large resistance $R$ across $110$ volt line, the meter reads $5V$ . Find the resistance $R$ in $k\Omega $ .

Answer 1

Hint: The current through each resistor in series is the same. So, find the current when the voltmeter is connected to the large resistance.
When two resistances are connected in series then its equivalent resistance is the algebraic sum of the two resistances.
Apply Ohm’s law to find the relation between the voltage and the current.

Complete step by step answer:
The rating of the voltmeter is given $150V,20k\Omega $ which the voltmeter has an internal resistance of $20k\Omega $ .
Now, if it is connected in series with a large resistance $R$ , the equivalent resistance will become $R + 20000$ as when two resistances are connected in series then its equivalent resistance is the algebraic sum of the two resistances.
It is given in the question that the meter reads $5V$ which means that the voltage across the $20k\Omega $ resistance becomes $5V$.
So, current across this resistance is given by $I = \dfrac{V}{R} = \dfrac{5}{{20000}}A$
We know that the current through each resistor in series is the same. This means that the current will be the same for the equivalent resistance and the circuit.
So, by applying Ohm’s law we have
$V = IR$
As given in the question that $V = 110V$ then
$110 = \dfrac{5}{{20000}}\left( {R + 20000} \right)$
On further solving we have
$\dfrac{5}{{20000}}R = 110 - 5 = 105$
On simplifying we have
$R = 20000 \times 21 = 420{\text{ k}}\Omega $ .
Hence, the resistance $R$ is $420{\text{ k}}\Omega $ .

Note: The Ohm’s law is valid only if the temperature and other physical conditions remain same as difference in temperature can cause change in value of the resistances.
A voltmeter is always connected in parallel to a resistance to find the voltage drop across it. It has a very high value of resistance. In fact, for an ideal voltmeter, the value of its internal resistance tends to infinity.