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

(ii) (a) Let I₁, I₂ and I₁ + I₂ be the currents flowing through the resistors r₁, r₂ and R respectively. Applying Kirchhoff’s law to the closed circuit CBAFC, we have
– 2 I₂ + 1 I₁ = 2.0 – 1.5 = 0
or, 2 I₂ – I₁ = 0.5 ...(i)
Again applying Kirchhoff’s law for closed circuit CFEDC, we have
– 1 I₁ – 5 (I₁ + I₂) + 1.5 = 0
or, 6 I₁ + 5 I₂ = 1.5 solving (i) and (ii), we get
                             
             
 Current through CF = 

Current through BA = 
 Current through DE = 
                      
(b) Potential difference across 5 Ω resistor