# Question #86804

Feb 27, 2017

(a) ${R}_{1} = 200 \Omega , {R}_{2} = 160 \Omega , {R}_{3} = 80 \Omega , \mathmr{and} {R}_{4} = 200 \Omega$
(b) $I = 110 \text{ mA}$

#### Explanation:

(a) We know that effective resistance of two resistors connected in series is sum of individual resistances
${R}_{s} = {R}_{1} + {R}_{2}$
and for two resistors connected in parallel is given by the expression
${R}_{p} = \frac{{R}_{1} {R}_{2}}{{R}_{1} + {R}_{2}}$

Keeping above in mind the effective resistance of the circuit is
${R}_{e} = {R}_{1} + \frac{\left({R}_{2} + {R}_{3}\right) {R}_{4}}{\left({R}_{2} + {R}_{3}\right) + {R}_{4}}$ ......(1)
Given is ${R}_{2} = 2 {R}_{3} , {R}_{1} = {R}_{4} , 2 {R}_{1} = 5 {R}_{3}$

Inserting in (1) above and making all in terms of ${R}_{3} \mathmr{and} {R}_{1}$
${R}_{e} = {R}_{1} + \frac{\left(2 {R}_{3} + {R}_{3}\right) {R}_{1}}{\left(2 {R}_{3} + {R}_{3}\right) + {R}_{1}}$
$\implies {R}_{e} = {R}_{1} + \frac{3 {R}_{3} {R}_{1}}{3 {R}_{3} + {R}_{1}}$
$\implies {R}_{e} = {R}_{1} + \frac{3 \left(\frac{2}{5} {R}_{1}\right) {R}_{1}}{3 \left(\frac{2}{5} {R}_{1}\right) + {R}_{1}}$
$\implies {R}_{e} = {R}_{1} + \frac{\frac{6}{5} {R}_{1}^{2}}{\frac{11}{5} {R}_{1}}$
$\implies {R}_{e} = {R}_{1} + \frac{6}{11} {R}_{1}$
$\implies {R}_{e} = \frac{17}{11} {R}_{1}$

Assuming that internal resistance of the voltage source is small and is ignored in the calculations
$I = \frac{V}{R}$
$\implies {I}_{1} = \frac{34}{R} _ e$
$\implies 0.11 = \frac{34}{\frac{17}{11} {R}_{1}}$
$\implies {R}_{1} = \frac{34 \times 11}{17 \times 0.11}$
$\implies {R}_{1} = 200 \Omega$
Now ${R}_{3} = \frac{2}{5} {R}_{1}$
$\implies {R}_{3} = \frac{2}{5} \times 200 = 80 \Omega$
Also ${R}_{2} = 2 {R}_{3}$
${R}_{2} = 2 \times 80 = 160 \Omega$
And ${R}_{1} = {R}_{4} = 200 \Omega$

(b) As seen from the circuit, both the Resistor ${R}_{1}$ and the voltage source are in series. In series circuit current is same in all the circuit components.
$\therefore I = {I}_{1} = 110 \text{ mA}$

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Text (11.9) [T/I] in the question is not relevant to the circuit.