A student connect a cell to a circuit and measures the current in the circuit as `I_1`. When he joins a second identical cell In seris with the first, the current becomes `I_2`. When he connects the cells in parallel, the current through the circuit is `I_3`. show that `3I_2I_3=2I_1(I_2+I_3)`.

Explanation

Suppse emf of each cell=E, internal resistance =r and resistance of the external circuit=R.
When there is only one cell in the circuit, `I_1=E/(R+r)`
When two cells are connected in series in the circuit
`I_2=(2E)/(R+2r)`
Again, when two cells are connected in parallel in the circuit,
`I_3=(2E)/(2R+r)`
Now, `1/I_2+1/I_3=(R+2r)/(2E)+(2R+r)/(2E)`
or,`(I_2+I_3)/(I_2I_3)=(3(R+r))/(2E)=3/2.(R+r)/E=3/2. 1/I_1`
or,`3I_2I_3=2I_1(I_2+I_3)` (Proved)