VI - From the ACGHA loop Potential difference at battery = - V₁
Potential difference at resistancess = + I₁ R₁ + (I₁
+ I₂ ) x R₃
The resultaint Potential difference in the loop is = - V₁
+ I ₁ R₁ + (I₁ + I₂ ) x R₃ = 0

Example 5 : Find the resultaint potential diffrence from the given figure based on Loops law.
Solution: Shown in For the loop ACDBA, -V₂ + I₂ R₂ – I₁ R₁ +V₁ = 0
For the loop EFDCE -(I₁ +I₂ ) R₃ – I₂ R₂ + V₂ = 0
For the loop EFBAE -(Ic₁ +I₂ ) R ₃ – I₁ R₁ +V₁ = 0


Example 6: Find electric current drawn(fig-30)from the battery of emf 12V.
Solution: Let I = I₁ +I₂ be the current drawn from emf 12V.
From the figure E. Using the loop law, For the loop DABCD, -3 (I₁ +I₂ ) + 12 – 5 – 2I₁ = 0       ……….. (a)
For the loop DAFED, -3 (I₁ +I₂ ) + 12 – 4I₂ = 0       ………. (b)
Solving the equation (a) & (b)
We get I₁ = 0.5 A and I₂ = 1.5 A


Total current drawn is then I = 0.5 + 1.5 = 2A
• You might have heard the sentences like “this month we have consumed 100 units of current”. What does ‘unit’ mean?
• A bulb is marked 60W and 120V. What do these values indicate? Let us see.

Electric power

The electric appliances that we use in our daily life like heater, cooker, fan, and refrigerator etc. consume electric energy.
Let us consider a conductor of resistance ‘R’ through which an electric current ‘I’ passes.
We know that when current passes through conductor, heat energy is generated.

Consider that a charge Q Coulomb passes through a point A, moves to point B