The cut-off voltage of the diodes (shown in figure) in forward bias is $$0.6$$ V. The current through the resister of $$40\Omega$$ is ______ mA.

Correct Answer: 4

Given:
Battery =1 V, diode drop =0.6 V

Step 1: Active path
Only one diode conducts ⇒ series path: 60Ω and 40Ω

Step 2: Apply KVL

1−i(60)−0.6−i(40)=0

1−0.6=100i

⇒0.4=100i

Step 3: Solve

i=$$\ \frac{\ 0.4}{100}$$ = 0.004 A

=4 mA

Final Answer:

4 mA

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