In a p-n junction diode under forward bias, the charge carriers responsible for current flow in the external circuit are free electrons. These electrons are present in the n-type material, and they move towards the p-type material. The free electrons from the n-side travel through the external circuit to the p-side, where they recombine with holes. This flow of free electrons constitutes the current in the connecting wire.
In a forward-biased p-n junction diode, the electrons from the n-side and the holes from the p-side move toward the junction. However, in the external connecting wire, the current is due to the movement of \(\textbf{free electrons}\), as electrons are the actual charge carriers in metal conductors (wires). Holes contribute to current only within the semiconductor, not in the external circuit.
The graph shows the variation of current with voltage for a p-n junction diode. Estimate the dynamic resistance of the diode at \( V = -0.6 \) V.

Assertion (A): We cannot form a p-n junction diode by taking a slab of a p-type semiconductor and physically joining it to another slab of an n-type semiconductor.
Reason (R): In a p-type semiconductor, \( n_e \gg n_h \) while in an n-type semiconductor \( n_h \gg n_e \).
The circuit shown in the figure contains two ideal diodes \( D_1 \) and \( D_2 \). If a cell of emf 3V and negligible internal resistance is connected as shown, then the current through \( 70 \, \Omega \) resistance (in amperes) is: 