Two cells of emf 2v and 4v and internal resistance 1 ohm and 2 ohm respectively are connected in parallel so as to send current in the same direction through an external resistance of 10 ohm.Find the potential difference across 10 ohm resistor.
Answers
The circuit diagram is shown in picture -
Applying Kirchoff's Law in loop ABCF -
2 = 10*I₁ + 10*I₂ + I₁
=> 2 = 11*I₁ + 10*I₂ ..........................(1)
Applying Kirchoff's Law in loop FCDE
4 - 2 = 2I₂ - I₁
=> 2 = 2I₂ - I₁ ............................(2)
Solving equations (1) and (2)
Multiplying equation (2) by 11 .
Thus, 2 = 11*I₁ + 10*I₂ and 22 = -11*I₁ + 22*I₂
=> 24 = 32I₂
=> I₂ = 24/32 = 3/4 A
So, 2 = 2*(3/4) - I₁
=> I₁ = (3/2) - 2
= - 1/2 A
So, I₁ + I₂ = -1/2 + 3/4
= 1/4 A
So, Potential Difference across 10 ohm resistor = V = I * R
= (1/4) * 10
= 2.5 V
=> Potential Difference across 10 ohm resistor = 2.5 V
