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) Let I1, I2 and I1 + I2 be the currents flowing through the resistors r1, r2 and R respectively.
On applying Kirchhoff’s law to the closed circuit CBAFC, we have
– 2 I2 + 1 I1 = 2.0 – 1.5 = 0
2 I2 – I1 = 0.5 ...(i)
Again, on applying Kirchhoff’s law for closed circuit CFEDC, we have
– 1 I1 – 5 (I1 + I2) + 1.5 = 0
6 I1 + 5 I2 = 1.5 ...(ii)
On solving (i) and (ii), we get
Current through CF,
Current through BA ,
Potential difference across 5 Ω resistor, V= IR
