Question

To register are connected in parallel then its net value is 3om and when connected in series then its net value is
6om. Find each resistance value.

Answer 1

Correct Question :

Two resistors are connected in parallel then its net value is 3 Ω and when connected in series then its net value is 16 Ω . Find each resistance value.

Given :

• Equivalent resistance in series circuit = 16 Ω

• Equivalent resistance in parallel circuit = 3 Ω

To find :

The value of each resistors

Solution :

We know the formula for Equivalent resistance i.e,

$\bf{R_{1} + R_{2} + R_{3} + .... = R_{n}}$

Where :

• R = Resistance

Here in our case two resistors are there , hence the formula will be :

$:\implies \bf{R_{1} + R_{2} = R_{n}}$

Now , by substituting the value of Rn in the above equation , we get :

$:\implies \bf{R_{1} + R_{2} = 16}$

Now, from this Equation , we get :

$:\implies \bf{R_{1} = 16 - R_{2}}$

$\boxed{\therefore \bf{R_{1} = 16 - R_{2}}}$ ⠀⠀⠀⠀⠀Eq.(i)

We know the formula for Equivalent resistance in case of a Parallel Circuit i.e,

$\bf{\dfrac{1}{R_{1}} + \dfrac{1}{R_{2}} + \dfrac{1}{R_{3}} + .... = \dfrac{1}{R_{n}}}$

Where :

• R = Resistance

Here in our case two resistors are there , hence the formula will be :

$\bf{\dfrac{1}{R_{1}} + \dfrac{1}{R_{2}} = \dfrac{1}{R_{n}}}$

Now , by substituting the value of Rn in the above equation , we get :

$\bf{\dfrac{1}{R_{1}} + \dfrac{1}{R_{2}} = \dfrac{1}{3}}$

By Substituting the value of R1 from eq.(i) , we get :

$\bf{\dfrac{1}{16 - R_{2}} + \dfrac{1}{R_{2}} = \dfrac{1}{3}}$

By solving it , we get :

$\bf{\dfrac{R_{2} + 16 - R_{2}}{16R_{2} - R_{2}^{2}} = \dfrac{1}{3}}$

$\bf{\dfrac{\not{R_{2}} + 16 - \not{R_{2}}}{16R_{2} - R_{2}^{2}} = \dfrac{1}{3}}$

$\bf{\dfrac{16}{16R_{2} - R_{2}^{2}} = \dfrac{1}{3}}$

By Cross-multiplication , we get :

$\bf{16 \times 3 = 16R_{2} - R_{2}^{2}}$

$\bf{48 = R_{2}^{2} - 16R_{2}}$

$\bf{48 = 16R_{2} - R_{2}^{2}}$

$\bf{0 = 16R_{2} - R_{2}^{2} - 48}$

$\bf{0 = - R_{2}^{2} + 16R_{2} - 48}$

$\bf{0 = - (R_{2}^{2} - 16R_{2} + 48)}$

$\bf{0 = R_{2}^{2} - 16R_{2} + 48}$

Using the quadratic equation formula and substituting the values in it, we get :

$\bf{\alpha = \dfrac{-b \pm \sqrt{b^{2} - 4ac}}{2a}}$

Here :

• a = 1
• b = (-16)
• c = 48

Let R1 be x .

$:\implies \bf{\alpha = \dfrac{-(-16) \pm \sqrt{(-16)^{2} - 4 \times 1 \times 48}}{2a}}$

$:\implies \bf{x = \dfrac{16 \pm \sqrt{256 - 192}}{2 \times 1}}$

$:\implies \bf{x = \dfrac{16 \pm \sqrt{64}}{2}}$

$:\implies \bf{x = \dfrac{16 \pm 8}{2}}$

$:\implies \bf{x = \dfrac{16 + 8}{2}\:/\:\dfrac{16 - 8}{2}}$

$:\implies \bf{x = \dfrac{24}{2}\:/\:\dfrac{8}{2}}$

$:\implies \bf{x = 12\:/\:4}$

$\boxed{\therefore \bf{R_{2}\:(x) = 12\:;\:4}}$

Hence the value of R2 is 12 or 4.

Hence, the R1 is 4 when R2 is 12 and R1 is 12 when R2 is 4.