tex]Two resistances when connected in parallel give resultant value of 2 Ω, when connected in series the value becomes 9Ω. Calculate

Let the value of each resistance be x and y.

Now, when x and y are in parallel, their equivalent resistance is 2 Ω

Therefore:

[tex]\tt\longrightarrow \dfrac{1}{2}=\dfrac{1}{x}+\dfrac{1}{y}[/tex]

[tex]\tt\longrightarrow \dfrac{1}{2}=\dfrac{x+y}{xy}[/tex]

[tex]\tt\longrightarrow xy=2(x+y) -(i)[/tex]

Now, when x and y are in series, their equivalent resistance is 9Ω

[tex]\tt\longrightarrow x+y =9-(iii)[/tex]

Substituting the value of x + y in (i), we get:

[tex]\tt\longrightarrow xy=2\times9[/tex]

[tex]\tt\longrightarrow xy=18[/tex]

[tex]\tt\longrightarrow x=\dfrac{18}{y}[/tex]

Substituting the value of x in (iii), we get:

[tex]\tt\longrightarrow y+\dfrac{18}{y}=9[/tex]

[tex]\tt\longrightarrow y^{2}+18=9y[/tex]

[tex]\tt\longrightarrow y^{2}-9y+18=0[/tex]

[tex]\tt\longrightarrow y^{2}-(6+3)y+18=0[/tex]

[tex]\tt\longrightarrow (y-6)(y-3)=0[/tex]

[tex]\tt\longrightarrow y=3, 6[/tex]

Therefore:

[tex]\tt\longrightarrow x=\dfrac{18}{3}, \dfrac{18}{6}[/tex]

[tex]\tt\longrightarrow x=6, 3[/tex]

So, the value of each resistance is 3 Ω and 6 Ω respectively.

Let R₁, R₂ and R₃ be 3 resistors connected in series. Let R be the equivalent resistance of the circuit.

Then:

→ R = R₁ + R₂ + R₃

If they were connected in parallel combination, then:

→ 1/R = 1/R₁ + 1/R₂ + 1/R₃

The SI unit of resistance is ohm. It is denoted by the Greek letter Ω (Omega)