Question

Two resistors are connected in series gives an equivalent resistance of 10 Ω. When connected in parallel, gives 2.4 Ω. What are the values of individual resistance , with complete solution​ , sorry I accidentally mentioned it as maths

Answer 1

Answer:

Resistors have resistance of 6Ω and 4Ω

Step-by-step explanation:

Let the Resistors be r1 and r2,

In Series,

Sum of Resistance in Series =

(Rs) = R1 + R2 + .........+ Rn

Now, According to the Question,

Here,

Rs = 10 Ω

Thus,

r1 + r2 = 10 ----- 1

also,

Sum of Resistance in Parallel =

(1/(Rp)) = (1/R1) + (1/R2) + ....... + (1/Rn)

Now, According to the Question,

Rp = 2.4 Ω ----- 2

Thus,

(1/r1) + (1/r2) = (1/Rp)

(r2 + r1)/(r1 × r2) = (1/Rp)

(r1 + r2)/(r1r2) = (1/Rp)

Thus, to get Rp we must take their Reciprocal,

Rp = (r1r2)/(r1 + r2)

Now, from eq.1 and eq.2 we get

2.4 = (r1r2)/10

r1r2 = 2.4 × 10

So,

r1 × r2 = 24

Thus, Now we are forced to stop but, No we still can continue,

Let's find the factors of 24

24 = 1, 2, 3, 4, 6, 8, 12, 24

Now, we can follow the above condition, they must add upto 10 and should multiply to get 24

So,

1 + 24 is not equal to 10

2 + 12 is not equal to 10

3 + 8 is not equal to 10

4 + 6 = 10

Thus, the individual resistors have resistance 6Ω and 4Ω

Remember, we don't need to check for the product because they are the factors of 24

So, the answer is 6Ω and 4Ω

Hope it helped and you understood it........All the best