KCET 2014 Physics Solved Paper

Let the two resistances be R1 and R2
When they are connected in series, equivalent resistance, Rs=6Ω
R1+R2=6 → (1)
When they are connected in parallel, equivalent resistance,
Rp=

4

3

Ω
‌

1

Rp

=‌

1

R1

+‌

1

R2

⇒‌

1

R1

+‌

1

R2

=‌

3

4

→ (2)
Solving Eq. (2), we get
⇒R1R2=6×‌

4

3

⇒R,R2=2×4=8Ω (Using Eq.(1))
⇒R1R2=2×4=8Ω → (3)
From Eq. (1), we have R2=6−R1
Using Eq. (3), we get R2=2 or 4
Therefore, the two resistances are 4 Ω and 2 Ω.