Thank you for reporting, we will resolve it shortly
Q.
The current flowing through $10\,\Omega$ resistor in the circuit shown in the figure is
Semiconductor Electronics: Materials Devices and Simple Circuits
Solution:
In the given circuit, junction diode $D_1$ is forward biased and will conduct whereas junction diode $D_2$ is reverse biased and will not conduct. Therefore current through $10\,\Omega$ resistor is
$I = \frac{2\,V}{10\,\Omega+15\,\Omega} = 0.08\,A = 80\,mA$
