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Q.
The current $ I $ through $ 10\,\Omega $ resistor in the circuit given below is
KEAMKEAM 2008Semiconductor Electronics: Materials Devices and Simple Circuits
Solution:
Here, diode $ {{D}_{2}} $ is reverse biased while $ {{D}_{1}} $ is forward biased. So, no current flows across $ {{D}_{2}}, $ current flows through diode $ {{D}_{1}} $
$ I=\frac{V}{R}=\frac{2}{10+15}=\frac{2}{25} $
$ =0.08A=80\,mA $
