Two cells of emf 1.5 V and 2 V and internal resistance 1`Omega` and 2`Omega` respectively are connected in parallel to pass a current in the same direction through an external resistance of 5`Omega`.
(i) Draw the circuit diagram.
(ii) Using Kirchhoff's laws, calculate the current through each branch of the circuit and potential difference across the 5`Omega` resistor.

Explanation

(i) The circuit diagram has been shown in Fig.

(ii) In mesh ACRDBA, applying Kirchhoff.s second law, we have
` - (I_1 + I_2) .5 - I_1 .1 + 1.5 = 0`
`rArr 6I_1 + 5I_2 = 1.5 ` ....(i)
and for mesh CRDC, we have
` - (I_1 + I_2) .5 - I_2 .2 + 2= 0`
` rArr 5I_1 + 7I_2 = 2` ....(ii)
On solving (i) and (ii), we get `I_1 = 1/34 A` and `I_2 = 9/34 A`
Current through external `(5Omega)` resistor
` = I_1 + I_2 = 1/34 + 9/34 = 10/34 A = 5/17 A`
` therefore ` Potential difference across the 5`Omega` resistor
`V = (I_1 + I_2) R = 5/17 xx 5 = 25/17 V`