Question

Find the equivalent resistance of the circuit. 400 20 a 80 w 30 W O a 85 1700 Ob. c. 80.50 d. 1000

Answer 1

For resistors connected in series,

Equivalent resistance,

Req R1 + R2 + R3+

and for resistors connected in parallel,

equivalent resistance, $\frac{1}{R_{eq}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}+.........$

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Consider the network

Resistors are connected in series

40Ω αnd 20Ω

$R_{s}=40\Omega +20\Omega$

${\color{Red} R_{s}=60\Omega }$

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Resistors $R_{s}\ and\ 30\Omega$ are connected in parallel

$\frac{1}{R_{p}}=\frac{1}{R_{s}}+\frac{1}{30\Omega }$

$\frac{1}{R_{p}}=\frac{1}{60\Omega }+\frac{1}{30\Omega }$

$\frac{1}{R_{p}}=\frac{1}{60}+\frac{1}{30}=\frac{1}{60}+\frac{2}{60}=\frac{1+2}{60}=\frac{3}{60}=\frac{1}{20}$

$\frac{1}{R_{p}}=\frac{1}{20}$

${\color{Red} R_{p}=20\Omega}$

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Resistors $80\Omega \ and\ R_{p}$ are connected in series

$R_{eq}=80\Omega +R_{P}$

$R_{eq}=80\Omega +20\Omega$

${\color{Red} R_{eq}=100\Omega }$

ANSWER: Option d. $100\ \Omega$

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