Question

Find the equivalent resistance of the circuit shown in the figure.



Assume that V = 21.6 V, R₁ = 3.44 Ω, R₂ = 4.67 Ω, R₃ = 4.11 Ω, R₄ = 8.05 Ω and R₅ = 4.08 Ω.

tv {R3

Answer 1

First R₂ and R₅ are in series

$R' = R_2 + R_ 5 = 4.67 \ \Omega + 4.08 \ \Omega = 8.75 \ \Omega$

Now R' and R₃ are in parallel so

$R'' = \frac{R'R_3}{R' + R_3}= \frac{8.75 \ \Omega \times 4.11 \ \Omega}{8.75 \ \Omega + 4.11 \ \Omega} = 2.80 \ \Omega$

Finally R'' , R₁ and R₄ are in series so the equivalent resistance of the circuit

$R_{eq} = R_1 +R''+ R_ 4 = 3.44 \ \Omega + 8.05 \ \Omega +2.80 \ \Omega$

$R_{eq} =14.29 \ \Omega$