Physics - Current Electricity Question with Solution | TestHub

Let $R$ be the resistance per unit length of the wires.

The length of wire AB is $L_{AB}=120cm$.

The length of wire CD is $L_{CD}=100cm$.

The resistance of wire AB is $R_{AB}=120R$.

The resistance of wire CD is $R_{CD}=100R$.

The emf of the ideal cell is $V=6V$.

The emf of the other cell is $E$, with internal resistance $r=2ohm$.

The ammeter reading is $I=2A$.

A length of $20cm$ of wire CD is balanced by $40cm$ length of wire AB.

This means the potential drop across $20cm$ of CD is equal to the potential drop across $40cm$ of AB.

Let $V_{CD}$ be the potential drop across $20cm$ of CD, and $V_{AB}$ be the potential drop across $40cm$ of AB.

So, $V_{CD}=V_{AB}$.

The resistance of $20cm$ of wire CD is $R_{CD,20}=20R$.

The resistance of $40cm$ of wire AB is $R_{AB,40}=40R$.

From the circuit, the current through wire AB is $I_{AB}$.

The total resistance in the main circuit (containing the $6V$ cell and wire AB) is $R_{total}=R_{AB}=120R$.

The current $I_{AB}$ is given by $I_{AB}=\frac{6V}{120R}=\frac{1}{20R}$.

The potential drop across $40cm$ of wire AB is $V_{AB}=I_{AB}\times R_{AB,40}=\frac{1}{20R}\times 40R=2V$.

Since $V_{CD}=V_{AB}$, the potential drop across $20cm$ of wire CD is $V_{CD}=2V$.

The current through wire CD is $I_{CD}$.

The potential drop $V_{CD}$ is also given by $V_{CD}=I_{CD}\times R_{CD,20}$.

So, $2V=I_{CD}\times 20R$.

This implies $I_{CD}=\frac{2}{20R}=\frac{1}{10R}$.

The ammeter reading is $2A$. This is the current flowing from the cell with emf $E$.

So, $I_{CD}=2A$.

Therefore, $\frac{1}{10R}=2A$.

This gives $10R=\frac{1}{2}$, so $R=\frac{1}{20}ohm/cm$.

Now we can find the emf $E$.

The current $I_{CD}$ is given by Ohm's law for the cell with internal resistance:

$I_{CD}=\frac{E}{R_{CD,total}+r}$.

Here, $R_{CD,total}$ is the resistance of the entire wire CD, which is $100R$.

$R_{CD,total}=100\times \frac{1}{20}=5ohm$.

The internal resistance $r=2ohm$.

So, $2A=\frac{E}{5ohm+2ohm}$.

$2A=\frac{E}{7ohm}$.

$E=2A\times 7ohm=14V$.

The emf $E$ is $14V$.