the effective resistacce of two resistors connected in series is 100 ohm.when it connected in parallel it is 24 ohm.determine the values of two resistors

R+r=100

1/R +1/r =1/24 => (R+r)/Rr=1/24

100/Rr =1/24
rR=2400

Since r=100-R
R(100-R)=2400
R²-100R+2400=0
R=40 Ω => r=60 Ω
R=60 Ω => r=40Ω

To determine the values of the resistors, let's use the formulas for calculating the effective resistance in series and parallel connections.

1. Series Connection:
In a series connection, the effective resistance is the sum of the resistances of the individual resistors.
Therefore, if the effective resistance in series is 100 ohms, it can be expressed as:
R1 + R2 = 100 (equation 1)

2. Parallel Connection:
In a parallel connection, the reciprocal of the effective resistance is equal to the sum of the reciprocals of the individual resistances.
Therefore, if the effective resistance in parallel is 24 ohms, it can be expressed as:
1/R1 + 1/R2 = 1/24 (equation 2)

Now, we have two equations with two unknowns (R1 and R2). We can solve these equations simultaneously to find the values of the resistors.

To do this, we can rearrange equation 2 to solve for one of the resistors:
1/R1 = 1/24 - 1/R2
1/R1 = (R2 - 24) / 24

Substituting this expression for 1/R1 in equation 1, we have:
(R2 - 24) / 24 + R2 = 100
(R2 - 24) + 24R2 = 2400
25R2 - 24 = 2400
25R2 = 2424
R2 = 2424 / 25
R2 = 96.96 ohms (approximately)

Now, we can substitute the value of R2 back into equation 1 to solve for R1:
R1 + 96.96 = 100
R1 = 100 - 96.96
R1 = 3.04 ohms (approximately)

Therefore, the values of the two resistors are approximately R1 = 3.04 ohms and R2 = 96.96 ohms.