Question

Two different resistors, if connected in series, give total resistance of 16 ohm and if connected in parallel, give total resistance of 3 ohm. Find each resistance.with sloution.

Answer 1

Answer

12 ohms and 4 ohms

Solution

Let the resistors be of resistance $\sf R_1$ and $\sf R_2$

When in series, the equivalent resistance is simple added up. Hence,

$\sf R_1 + R_2 = 16$

For parallel connection, the sum of inverse of resistance is equal to the inverse of equivalent resistance. Hence,

$\sf \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{3}$

$\sf \frac{R_1 + R_2}{R_1R_2} = \frac{1}{3}$

We know that $\sf R_1 + R_2 = 16$

Hence,

$\sf \frac{16}{R_1R_2} = \frac{1}{3}$

$\sf\implies R_1R_2 = 48$

Now, by hit and trial we can say that the value of resistances is 12 and 4 ohm, but here is the longer method

$\sf (R_1 + R_2)^2 = (16)^2$

$\sf (R_1)^2 + (R_2)^2 + 2R_1R_2 = 256$

Subtract $\sf 4R_1R_2$ from both sides

$\sf (R_1)^2 + (R_2)^2 - 2R_1R_2 = 256 - 4R_1R_2$

$\sf (R_1 - R_2)^2 = 256 - 4(48)$

$\sf (R_1 - R_2)^2 = 64$

Taking square root on both sides,

$\sf R_1 - R_2 = 8$

We know that $\sf R_1 + R_2 = 16$

Adding the two equations, we get

$\sf R_1 + R_2 + R_1 - R_2 = 16 + 8$

$\sf 2R_1 = 24$

$\sf R_1 = 12 \ ohm$

Therefore, $\sf R_2 = 4\ ohm$

Answer 2

Answer:

${\boxed{\bold\green{12\:ohm\: \: and \: \: 4 \:ohm}}}$

Step-by-step explanation:

In a series, two resistors give total resistance of 16 ohm. If connected in parallel, it gives total resistance of 3 ohm.

$\sf At\:first, R_1 + R_2 = 16 .........(i)$

$\sf Secondly, \dfrac{1}{R_1} + \dfrac{1}{R_2} = \dfrac{1}{3}$

→ $\sf \dfrac{R_2+R_1}{R_1 R_2} = \dfrac{1}{3}$

$\sf *From\:(i)\:we\:get:-$

→ $\sf \dfrac{16}{R_1 R_2} = \dfrac{1}{3}$

→ $\sf R_1 R_2 = 48............(ii)$

$\rule{300}{1}$

$\tt *From\:(i)\::-$

→ $\sf {(R_1 + R_2)}^{2} = {16}^{2}$

↓ $\sf {(R_1 - R_2)}^{2} + 4R_1 R_2 = 256 \: \: \: [\because {(a+b)}^{2} ={(a-b)}^{2} + 4ab]$

→ $\sf {(R_1 - R_2)}^{2} + 4\times 48 = 256$

→ $\sf {(R_1- R_2)}^{2} = 256 - 192$

→ $\sf {(R_1 - R_2)}^{2} = 64$

→ $\sf (R_1 - R_2) =\sqrt{64}$

→ $\sf R_1 - R_2 = 8 .......(iii)$

$\tt Adding \: (i) \: and \: (iii)\: we\:get:-$

$\sf R_1 +\cancel{ R_2} + R_1 -\cancel{ R_2} = 16 + 8$

→ $\sf 2R_1 = 24$

→ $\sf R_1 =\dfrac{24}{2}$

→ ${\boxed{\sf{R_1 = 12\:ohm}}}$

$\sf R_2 = 16 - R_1$

→ $\sf R_2 = 16 - 12$

→ ${\boxed{\sf{R_2 = 4\:ohm}}}$