• # question_answer The current through $10\Omega$ resistor in the circuit given below is (the diodes are ideal) A)  50 mA                        B)  20 mAC)  40 mA                        D)  80 mA

${{D}_{2}}$is reverse biased. As it is an ideal diode, its resistance is infinite and no current flows through it, ${{D}_{1}}$is forward biased. As it is an ideal diode, its resistance is zero and hence, the current through $10\Omega$resistor is $I=\frac{2}{10+15}=0.08\,A=80\,mA$ Hence, the correction option is [d].