Question

Q 5 To get 2 Ω resistance using only 6 Ω resistors, the number of them required is *

Answer 1

Answer:

The answer to your question is 3.

Answer 2

Answer:

The number of them required is 3.

Explanation:

The resistance in circuits that obey Ohm's Law $V=IR$are about current flow.The total current flows into every resistor is series. However, only part of the current flows into each resistor in parallel.

The resistors in parallel only block part of the current Ohm's Law has the form $I = \frac{V }{ R}$for each resistor and partial current.

$I=I_1+I_2$

$I_1+I_2= \frac{V_1}{ R_1} + \frac{V_2}{ R_2}$

$I = \frac{V }{ R} = \frac{V}{ R_1} + \frac{V}{R_2}$

$R_{eq}= {( \frac{1}{R_1} + \frac{1}{R_2} ) }^{ - 1}$

Here resultant resistance is less than each resistance.

Therefore there resistance are to be connected in parallel combination.

Let, the number of them be n.

We know,

$R = {( \frac{1}{R_1} + \frac{1}{R_2} + ........ + \frac{1}{R_n} ) }^{ - 1}$

Here,

$R_1 = R_2 = ...... = R_n = R = 6$Ohm.

$R_{eq} = {( \frac{1}{R} + \frac{1}{R} + ........ + \frac{1}{R} ) }^{ - 1}$

$R _{eq}= \frac{1}{n \times \frac{1}{R} } = \frac{1}{n} \times R$

$R_{eq} = 2$and $R = 6$

So,

$2 = \frac{1}{n} \times 6 \\ n = 3$