Question

The ratio of two resistors A and B connected in series is $1:4$ and the current passing through them is $10A$. Then the ratio of the current that flows through them and when connected across the same potential difference is
$A)\text{ }4:1$
$B)\text{ 1}:4$
$C)\text{ 1}:2$
$D)\text{ 2}:1$

Answer 1

Hint: A Quincke’s tube is a device which is used to demonstrate the interference effects in standing sound waves. This device consists of a resonance tube with a millimetre scale, is partially filled with water and is connected to an expansion vessel with a tube.

This problem can be solved using the fact that for the same potential difference, the current through a resistor is inversely proportional to the value of the resistance. Therefore, the ratio of the currents in the two resistors can be found out from the ratio of the resistances.

Formula used:
$I\propto \dfrac{1}{R}$

Complete step by step answer:
For the same potential difference across a resistor of resistance $R$ and current $I$ passing through it, the relation is given by
$I\propto \dfrac{1}{R}$ --(1)
Now, let us analyze the question.
Let the resistance of the resistor A be ${{R}_{A}}$ and that of resistor B be ${{R}_{B}}$.
It is given that ${{R}_{A}}:{{R}_{B}}=1:4$.
$\therefore \dfrac{{{R}_{A}}}{{{R}_{B}}}=\dfrac{1}{4}$ --(2)
When connected across the same potential difference, let the currents in the resistors A and B be ${{I}_{A}}$ and ${{I}_{B}}$.
Therefore, using (1), we get
$\dfrac{{{I}_{A}}}{{{I}_{B}}}=\dfrac{{{R}_{B}}}{{{R}_{A}}}$
Putting (2) in the above equation, we get
$\dfrac{{{I}_{A}}}{{{I}_{B}}}=\dfrac{4}{1}=4:1$
$\therefore {{I}_{A}}:{{I}_{B}}=4:1$
Hence, the ratio of the currents in the resistors A and B is $4:1$.
Therefore, the correct option is $A)\text{ }4:1$.


Note:
Students must not get confused upon seeing that the value of the current in the question is not being used anywhere in the calculation. The current value would be of importance if the potential difference across each resistor when combined in series had to be calculated to get the total potential drop across the combination. However, here it is already mentioned that the same potential drop is applied across both resistors and hence, finding out the individual potential drops in the series combination is not necessary.