When two resistors of resistance R1 and R2 are connected in parallel, ...
Given:
Resistance of first resistor, R1 = ?
Resistance of second resistor, R2 = ?
Net resistance in parallel, Rp = 3 Ω
Net resistance in series, Rs = 16 Ω
Solution:
Let's use the formulas for calculating the net resistance of resistors in parallel and series.
For parallel connection:
The net resistance for two resistors in parallel is given by:
1/Rp = 1/R1 + 1/R2
Substituting the given value of Rp as 3 Ω, we get:
1/3 = 1/R1 + 1/R2
Equation (1)
For series connection:
The net resistance for two resistors in series is given by:
Rs = R1 + R2
Substituting the given value of Rs as 16 Ω, we get:
16 = R1 + R2
Equation (2)
Solving the equations:
We have two equations (1) and (2) with two variables R1 and R2. We can solve these equations simultaneously to get the values of R1 and R2.
From equation (2), we can solve for R2:
R2 = 16 - R1
Substituting this value of R2 in equation (1), we get:
1/3 = 1/R1 + 1/(16 - R1)
Multiplying both sides by 48R1(16 - R1), we get:
16R1 - R1^2 = 48 + 3R1(16 - R1)
Simplifying and rearranging, we get a quadratic equation:
R1^2 - 13R1 + 48 = 0
Factoring the equation, we get:
(R1 - 8)(R1 - 5) = 0
So, the possible values of R1 are 8 Ω and 5 Ω.
Final answer:
Therefore, the values of R1 and R2 are:
R1 = 8 Ω and R2 = 16 - R1 = 8 Ω
OR
R1 = 5 Ω and R2 = 16 - R1 = 11 Ω
So, the two resistors can have values of 8 Ω and 8 Ω or 5 Ω and 11 Ω to satisfy the given conditions.