Two ideal diodes are connected in the network as shown in figure. The equivalent resistance between A and B is __________ Ω.

Solution

Physics sir · 4 steps

Decide which diode conducts first. Then the two 20 Ω resistors fall into parallel.

1. Left diode. A is at the higher potential. The left diode points from A's node to the top node, so it is forward biased. An ideal diode that conducts acts as a plain wire, so the top node is at the same potential as A.
2. Right diode. The right diode points from B's node towards the top node. The top node is at A's potential, which is higher than B's, so this diode is reverse biased. It acts as an open switch and drops out of the circuit.
3. Redraw the network. With the top node joined to A, the middle 20 Ω and the lower-left 20 Ω both run from A to the bottom node, so they are in parallel. The 15 Ω then carries the current from the bottom node to B, in series with that pair.

   R_(AB) = (20 × 20)/(20 + 20) + 15 = 10 + 15 = 25 Ω

4. So. The equivalent resistance between A and B is 25 Ω.

The common slip: 35. Treating the left diode as reverse biased leaves only the 20 Ω and 15 Ω in series, which gives 35, not 25.

The idea to keep. With ideal diodes, first use the given polarity to decide which diode is on and which is off. Replace an on diode with a wire and an off diode with a break, then solve the plain resistor network.

Answer from the NTA final answer key. Solution by PrepWiser.