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

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, $I_1 = \frac{5}{100}$
$= 0.05\, A$
According to Kirchhoff’s, current rule,
$I_1 = I_2 + I_3 + I_4$ or
$I_2 + I_4 = I_1 = 0.05 \,A$
$\because R_{AB} = R_{EF}$, so, $I_{4} = I_{2}$
$ 2I_{4} = 2I_{2} = 0.05 ;$
$I_{4} = I_{4} = \frac{0.05}{2} $
$ = 0.025\, A$