Two ideal diodes are connected in the network as shown in figure. The equivalent resistance between $$A$$ and $$B$$ is ______ $$\Omega$$.

We need to find the equivalent resistance between terminals $$A$$ and $$B$$ for the given diode-resistor network.

1. Analyze the Diode Conditions

From the schematic shown , terminal $$A$$ is connected to the positive terminal ($$+$$) of the voltage source, and terminal $$B$$ is connected to the negative terminal ($$-$$).

Let's determine the bias state of each ideal diode based on the conventional current flowing from $$A$$ to $$B$$:

• Left Diode: The current entering from terminal $$A$$ encounters the p-side (anode) of the left diode. This makes it forward-biased, meaning it acts as a short circuit (closed switch or simple connecting wire).
• Right Diode: The current attempting to reach terminal $$B$$ encounters the n-side (cathode) of the right diode. This makes it reverse-biased, meaning it acts as an open circuit (broken wire or infinite resistance), and no current can pass through this upper-right branch.

2. Simplify the Circuit

By substituting the forward-biased diode with a short circuit and removing the reverse-biased branch completely, the network simplifies significantly:

• The top junction is now directly tied to the left junction (terminal $$A$$) via the shorted left diode.
• This places the upper central vertical $$20\ \Omega$$ resistor directly in parallel with the lower-left diagonal $$20\ \Omega$$ resistor.

Let's calculate the equivalent resistance ($$R_p$$) of these two parallel $$20\ \Omega$$ resistors:

$$R_p = \frac{20 \times 20}{20 + 20} = \frac{400}{40} = 10\ \Omega$$

3. Calculate the Total Equivalent Resistance ($$R_{AB}$$)

This parallel combination ($$R_p = 10\ \Omega$$) is in series with the remaining lower-right diagonal $$15\ \Omega$$ resistor leading straight to terminal $$B$$:

$$R_{AB} = R_p + 15\ \Omega$$

$$R_{AB} = 10\ \Omega + 15\ \Omega = 25\ \Omega$$

Conclusion

The equivalent resistance between $$A$$ and $$B$$ is 25 $$\Omega$$.