The value of net resistance of the network as shown in the given figure is :

Solution

Physics sir · 3 steps

Before adding anything, check which diodes conduct; a reverse-biased branch drops out completely.

1. Find current direction. The left end is at -6 V and the right end is at -8 V. Since -6 > -8, the left end is at the higher potential, so current tries to flow from left to right.
2. Check each diode. In the middle branch the diode's arrow points right, along the current. It is forward biased and acts like a plain wire. In the bottom branch the arrow points left, against the current. It is reverse biased, so that branch carries no current.

   Branch	Diode	Conducts?
   15 Ω	none	yes
   10 Ω	points right, forward biased	yes
   5 Ω	points left, reverse biased	no, open circuit

   Rₙₑₜ = (15 × 10)/(15 + 10) = 150/25 = 6 Ω

3. So. Only the 15 Ω and 10 Ω branches are in parallel. Together they give a net resistance of 6 Ω, option (b).

The idea to keep. In any circuit with ideal diodes, first find which end is at the higher potential. A forward-biased diode is a wire, and a reverse-biased diode cuts its whole branch out.

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