Question

The equivalent resistance of the circuit across $AB$ is given by (the diodes are ideal)

• A. $6\Omega$ or $9\Omega$
• B. $4\Omega$ or $9\Omega$
• C. $4\Omega$ or $13\Omega$
• D. $6\Omega$ or $0\Omega$

Answer 1

The correct option is C $4\Omega$ or $13\Omega$

For forward biased condition as shown
both the diode works as forward bias, so network works as a Wheatstone bridge.
hence, no currect flows from $5\Omega$ resistor.

hence, $R_{eq}=\frac{6\times 12}{6+12}=4\Omega$

For reverse biased condition as shown

Both the diode works as a open circuit.
hence, $R_{eq}=4+5+4=13\Omega$

Hence option C is correct.