The circuit shown in following figure contains two diode D1 and D2 each with a forward resistance of 50 ohms and with in finite backward resistance. If the battery voltage is 6 V, the current through the 100 ohm resistance (in amperes) is

Question

The circuit shown in following figure contains two diode D1 and D2 each with a forward resistance of 50 ohms and with in finite backward resistance. If the battery voltage is 6 V, the current through the 100 ohm resistance (in amperes) is (A) zero (B) 0.02 (C) 0.03 (D) 0.036

Tardigrade Admin · Accepted Answer

According to the given polarity, diode D1 is forward biased while D2 is reverse biased. Hence current will pass through D1 only. So current i=(6/(150+50+100))=0.02 A

Q. The circuit shown in following figure contains two diode $D_{1}$ and $D_{2}$ each with a forward resistance of $50 o hm s$ and with in finite backward resistance. If the battery voltage is $6 V$ , the current through the $100 o hm$ resistance (in amperes) is

zero

0.02

0.03

0.036

According to the given polarity, diode $D_{1}$ is forward biased while $D_{2}$ is reverse biased. Hence current will pass through $D_{1}$ only.
So current $i = \frac{6}{( 150 + 50 + 100 )} = 0.02 A$

Q. The circuit shown in following figure contains two diode D1​ and D2​ each with a forward resistance of 50ohms and with in finite backward resistance. If the battery voltage is 6V, the current through the 100ohm resistance (in amperes) is