Question

If each diode in figure has a forward bias resistance of $25\Omega$ and infinite resistance in reverse bias, The values of the current

• A. $I_2=0.40A,I_4=0.025A$
• B. $I_2=0.25A,I_4=0.20A$
• C. $I_1=0.05A,I_3=0.02A$
• D. $I_2=I_4=0.025A$

Answer 1

Answer: (D)

Let $R$ be the effective resistance of the circuit, then
$R=\left(R_{AB}\left|\right|R_{EF}\right)+25$
$R_{AB}=125+25=150\Omega ;$
$=125+25=150\Omega$
$\therefore R=25+\frac{150}{2}$
$=100\Omega$
Since diode in the branch $CD$ is reverse biased,
$I_3=0$.
Current, $I_1=\frac{5}{100}$
$=0.05A$
According to Kirchhoff’s, current rule,
$I_1=I_2+I_3+I_4$ or
$I_2+I_4=I_1=0.05A$
$\because R_{AB}=R_{EF}$, so, $I_4=I_2$
$2I_4=2I_2=0.05;$
$I_4=I_4=\frac{0.05}{2}$
$=0.025A$