Question

An enquiring physics student connects a cell to a circuit and measures the current drawn from the cell to $I_1$. When he joins a second identical cell in series with the first, the current becomes $I_2$. When the cells are connected in parallel, the current through the circuit is $I_3$. Show that relation between the current is $3I_3{I}_2=2I_1(I_2+I_3)$.

Answer 1

let the voltages of the cell be V and R the equivalent resistance of the circuit.

$I_1=\frac{V}{R}\because$ only one cell connected ;
$I_2=\frac{V+V}{R}=\frac{2V}{R}\because$ two cells connected in series ;
$I_3=\frac{V}{R}+\frac{V}{R}=\frac{2V}{R}\because$ two cells connected in parallel $\Rightarrow$ total current is same as sum of the currents due to each.
Now ,
LHS = $3I_3{I}_2=3\frac{2V}{R}\frac{2V}{R}=3\left(\frac{4V^2}{R^2}\right)$
RHS = $2I_3(I_1+I_2)=2\frac{2V}{R}(\frac{V}{R}+\frac{2V}{R})=2\frac{2V}{R}\frac{3V}{R}=3\frac{4V^2}{R^2}$
therefore, LHS = RHS