In given network, the resultant resistance between
A and B
Answers
Answer
Req = R/2
Refer to the attachment
More Information About Series Combination and parallel Combination
In series Combination
i) Current (i) is same for all resistors
ii) Potential Drop(v) is Distributed v=v₁ + v₂ + v₃
v=v₁ + v₂ + v₃
IRe = IR₁ + IR₂ + IR₃
IRe = I(R₁+R₂+R₃)
Re = R₁+R₂+R₃
In Parallel Combination
i) Potential (v) is same in all resistors
ii)Current(i) is distributed
I = I₁ + I₂ + I₃
V/Re = V/R₁ + V/R₂ + V/R₂
1/Re = 1/R₁+1/R₂+1/R₃
Explanation:
Given
⇒Tanθ = 20/21
To Find
⇒(1-Sinθ + Cosθ)/(1+Sinθ+Cosθ)
First of all We have to find Sinθ and Cosθ
So take
⇒Tanθ = 20/21 = Perpendicular(p)/Base(b)
We get
⇒Perpendicular = 20 , Base(b) = 21 and Hypotenuse(h) = h
Using Pythagoras theorem
⇒h² = p² + b²
⇒h² = (20)² + (21)²
⇒h² = 400 + 441
⇒h² = 841
⇒h = √(841)
⇒h = 29
We get
⇒Perpendicular = 20 , Base(b) = 21 and Hypotenuse(h) = 29
Then
⇒Sinθ = P/h and Cosθ = b/h
⇒Sinθ = 20/29 and Cosθ = 21/29
Now Put the value on
⇒(1-Sinθ + Cosθ)/(1+Sinθ+Cosθ)
⇒(1-20/29 + 21/29)/(1+20/29 + 21/29)
⇒{(29-20+21)/29}/{29+20+21)/29}
⇒{(50 - 20)/29}/{(50+20)/29}
⇒(30/29)/(70/29)
⇒30/29 ×29/70
⇒30/70
⇒3/7
Answer = 3/7
Answer
Req = R/2
Refer to the attachment
More Information About Series Combination and parallel Combination
In series Combination
i) Current (i) is same for all resistors
ii) Potential Drop(v) is Distributed v=v₁ + v₂ + v₃
v=v₁ + v₂ + v₃
IRe = IR₁ + IR₂ + IR₃
IRe = I(R₁+R₂+R₃)
Re = R₁+R₂+R₃
In Parallel Combination
i) Potential (v) is same in all resistors
ii)Current(i) is distributed
I = I₁ + I₂ + I₃
V/Re = V/R₁ + V/R₂ + V/R₂
1/Re = 1/R₁+1/R₂+1/R₃