Question

Circuit given in Fig(i) is equivalent to circuit given in Fig (ii) then

• A. $E_{eq}=3V$
• B. $r_{eq}=1\Omega$
• C. $E_{eq}=17V$
• D. Current in $1\Omega$ flow from B to A

Answer 1

Answer: (B), (C)

$E_{eq}=17V$

$\frac{1}{r_{eq}}=\frac{1}{2}+\frac{1}{6}+\frac{1}{3}\Rightarrow r_{eq}=1\Omega$
$E_{eq}=(\frac{20}{2}+\frac{18}{6}+\frac{12}{3})r_{eq}=17V$
and A is at higher potential then B.