Question

Two resistors are connected in series with 5V battery of negligible into nil resistance. A current of 2 A flows through each resistor. If they, are connected in parallel with the same battery a current of \(\frac{25}{3}\) Allows through combination. Calculate the value of each resistance.

Answer 1

Given: 

R₁ = ?, R₂ = ?

In series: R₁ + R₂ = R_S 

R_s = \(\frac{V}{I}=\frac{5}{2}\) = 2.5Ω 

R₁ + R₂ = R_s = 2.5Ω ….. (1) 

In parallel:

R_p = 0.6Ω

R_p = \(\frac{R_1R_2}{R_1+R_2}\)

From eq (1)

R_p = \(\frac{R_1R_2}{2.5}\)

= 0.6 × 2.5 = R₁R₂ 

R₁R₂ = 1.5Ω 

(R₁ – R₂)² = (R₁ + R₂)² – 4R₁R₂ 

= (2.5)² – 4 × 1.5 

= 6.25 – 6 

R₁ – R₂ = √0.25 

R₁ – R₂ = 0.5 ………. (2) 

Solve (1) and (2) 

R₁ + R₂ = 2.5 

R₁ – R₂ = 2.5 

2R₁ = 3 

R₁ = \(\frac{3}{2}\) = 1.5Ω 

From (1), we can write, 

R₁ + R₂ = 2.5 

1.5 + R₂ = 2.5 

R₂ = 2.5 -1.5 

R₂ = 1Ω

Comments

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