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Answers
Whenever we are given circuit containing more than one battery, it is better to apply KVL to find current across each resistors
For Loop - 1 [Anti-clockwise] :
→ 80 - 20I + 20 - 20I₁ = 0
→ 100 - 20I - 20I₁ = 0
→ 20 (I + I₁) = 100
→ I + I₁ = 5
→ I₁ = 5 - I _______ [Eq. 1]
For Loop - 2 [Clockwise] :
→ 20 - 20I₁ + 30I₂ = 0
→ 10 (2I₁ - 3I₂) = 20
→ 2I₁ - 3I₂ = 2
→ 2I₁ - 3 (I - I₁) = 2 (∵ I = I₁ + I₂)
→ 2I₁ - 3I + 3I₁ = 2
→ -3I + 5I₁ = 2 _______ [Eq. 2]
By comparing both equations,
→ -3I + 5 (5 - I) = 2
→ -3I + 25 - 5I = 2
→ -8I = -23
→ I = 23/8 A
By substituting the value of I into first equation,
→ I₁ = 5 - 23/8
→ I₁ = (40 - 23) / 8
→ I₁ = 17/8 A
We know that, I = I₁ + I₂
→ 23/8 - 17/8 = I₂
→ I₂ = 6/8
→ I₂ = 3/4 A
A] Current through 20Ω (I) = 23/8 A
B] Current through 30Ω (I₂) = 3/4 A
C] Current through 20Ω (I₁) = 17/8 A
Answer:
Whenever we are given circuit containing more than one battery, it is better to apply KVL to find current across each resistors..
For Loop - 1 [Anti-clockwise] :
→ 80 - 20I + 20 - 20I₁ = 0
→ 100 - 20I - 20I₁ = 0
→ 20 (I + I₁) = 100
→ I + I₁ = 5
→ I₁ = 5 - I _______ [Eq. 1]
For Loop - 2 [Clockwise] :
→ 20 - 20I₁ + 30I₂ = 0
→ 10 (2I₁ - 3I₂) = 20
→ 2I₁ - 3I₂ = 2
→ 2I₁ - 3 (I - I₁) = 2 (∵ I = I₁ + I₂)
→ 2I₁ - 3I + 3I₁ = 2
→ -3I + 5I₁ = 2 _______ [Eq. 2]
By comparing both equations,
→ -3I + 5 (5 - I) = 2
→ -3I + 25 - 5I = 2
→ -8I = -23
→ I = 23/8 A
By substituting the value of I into first equation,
→ I₁ = 5 - 23/8
→ I₁ = (40 - 23) / 8
→ I₁ = 17/8 A
We know that, I = I₁ + I₂
→ 23/8 - 17/8 = I₂
→ I₂ = 6/8
→ I₂ = 3/4 A
A] Current through 20Ω (I) = 23/8 A
B] Current through 30Ω (I₂) = 3/4 A
C] Current through 20Ω (I₁) = 17/8 A
