Answer to Question #153685 in Electric Circuits for cab

Question

Consider the following circuit. If the following are the corresponding values of the resistive elements indicated and 15V is applied,

1. What is the total equivalent resistance that the voltage source “sees”?
2. What is the current, I? Express answer in amperes. 

Please see attached image: https://ibb.co/3rSHhm3

Answer 1

(a) Refer to the figure,

Since, R₁₃ and R₁₄ are in series , r₁ = 8+5 = 13$\Omega$

This r₁ is in parallel to R₁₁ and R₁₂, so $\frac{1}{r_2} = \frac{1}{4}+\frac{1}{7}+\frac{1}{13} \implies r_2 = 2.13 \Omega$

r₂ is in series with R₉ and R ₁₀, then r₃ = 2.13 + 8+1 = 11.13 $\Omega$

This r₃ is in parallel to R_8, then $\frac{1}{r_4}=\frac{1}{8}+\frac{1}{11.3} \implies r_4 = 4.68 \Omega$

r₄ is in series with R₆ R₇, then r₅ = 4.68 + 6+3 = 13.68 $\Omega$

R₄ R₅ are in series, then r₆ = 1+7=8 $\Omega$

Now, r₅ r₆ R₂ and R₃ are in parallel, then $\frac{1}{r_7} = \frac{1}{1}+\frac{1}{6}+\frac{1}{8}+\frac{1}{13.68} \implies r_7 = 0.73\Omega$

Finally, r₇ and R₁ are in series, R = 1 + 0.73 = 1.73 $\Omega$

So equivalent resistance of the circuit is $R =1.73 \Omega$

(b) Current in the circuit is given by,

$I = \frac{V}{R} = \frac{15}{1.73} = 8.67 A$