Question

In the circuit of figure, treat diode as ideal, current in the $4\Omega$ resistor is

• A. $2A$
• B. $3A$
• C. $\frac{12}{7}A$
• D. $\frac{30}{13}A$

Answer 1

Answer: (D)

$\frac{30}{13}A$

Since both the diodes are forward biased, thus both acts as short circuited.
Equivalent circuit is shown above.
$2\Omega$ and $3\Omega$ are connected in parallel to each other and this combination is connected in series with $4\Omega$.
Equivalent resistance of the circuit $R_{eq}=(2\parallel 3)+4$
$R_{eq}=\frac{2\times 3}{2+3}+4=\frac{26}{5}\Omega$
Current through $4\Omega$, $I=\frac{V}{R_{eq}}$
$⟹R_{eq}=\frac{12}{26/5}=\frac{30}{13}A$