Two resistance when connected in parallel ,give resultant value of 2 ohm. When connected in series and value becomes 9 ohm.Calculate the value of each resistance.
Answers
Let's denote the first resistor by R1 and second resistor by R2.
Two resistors when connected in parallel, give resultant value of 2 ohm. (Rp = 2 ohm)
1/Rp = 1/R1 + 1/R2
1/2 = 1/R1 + 1/R2
1/2 = (R1 + R2)/R1R2 .........(1st equation)
Two resistors when connected in series and value becomes 9 ohm. (Rs = 9 ohm)
Rs = R1 + R2
9 = R1 + R2 ........................(2nd equation)
Substitute value of (2nd equation) in (1st equation)
1/2 = 9/R1R2
Also, we can R1 as (9-R2)
1/2 = 9/R2(9-R2)
1/2 = 9/(9R2 - (R2)²)
9R2 - (R2)² = 18
(R2)² - 9R2 + 18 = 0
(R2)² - 6R2 - 3R2 + 18 = 0
R2(R2 - 6) -3(R2 - 6) = 0
(R2 - 3) (R2 - 6) = 0
R2 = 3, 6
R1 = 9-3, 9-6 = 6, 3
Therefore, if R2 is 3 ohm then R1 is 6 ohm and if R2 is 6 ohm then R1 is 3 ohm.
GIVEN:
- Two resistors are connected in parallel give resultant value ( Rp ) = 2Ω
- When connected in series it becomes 9Ω
TO FIND:
- Value of each resistance
SOLUTION:
Let
- Resistors be R1 & R2
Given, two resistors when connected in series become (Rs) = 9Ω
That means,
Rs = R1 + R2
R1 + R2 = 9Ω
_____________
Two resistors when connected in parallel become = 2Ω
1/Rp = 1/R1 + 1/R2
→ 1/Rp = (R1 + R2)/R1×R2
Substituting the values R1 + R2 = 9 & Rp = 2
→ 1/2 = 9/R1 × R2
Cross multiply
→ 18Ω = R1 × R2
Substituting R1 = 9 - R2
→ (9 - R2)R2 = 18Ω
→ 9R2 - (R2)² = 18Ω
→ (R2)² - 9R2 + 18 = 0
Now we got a quadratic equation . Finding the value of R2 by factorisation
→ (R2)² - 6R2 - 3R2 + 18 = 0
Taking common
→ R2(R2 - 6) - 3(R2 - 6) = 0
→ (R2 - 6)(R2 - 3) = 0
R2 = (6 or 3)Ω
Finding R1
R1 = 9 - R2
Taking R2 = 6
R1 = 9 - 6
R1 = 3Ω
Taking R2 = 3
R1 = 9 - 3
R1 = 6Ω
Hence, if R1 = 3Ω , R2 = 6Ω & if R1 = 6Ω , R2 = 3Ω , that means
R1 = 6Ω or 3Ω
R2 = 3Ω or 6Ω