Series diodes carry identical current: Two diodes are connected in series in the same branch. The first diode has a forward voltage of 0.75 V and the second has 0.80 V. If the current through the first diode is 500 mA, what is the current through the second diode?

Introduction / Context:In any series branch, the same current flows through all components because there is only one path for charge. This applies to resistors, diodes, LEDs, and other two-terminal devices. Forward voltage drops may differ between devices, but that does not change the equality of current in series.

Given Data / Assumptions:

• Two diodes in series in one branch.
• Measured/assumed forward drops: 0.75 V and 0.80 V.
• Measured current through the first diode: 500 mA.

Concept / Approach:Series-circuit rule: the same current passes sequentially through each element in the branch. The supply must provide a voltage equal to the sum of all drops (including any series resistor), but the current is identical through both diodes at any instant in steady operation.

Step-by-Step Solution:Identify topology: single series path → same current everywhere in that path.Measured I through diode 1 = 500 mA.Thus I through diode 2 = 500 mA.Total forward drop across both diodes = 0.75 V + 0.80 V = 1.55 V (for completeness).

Verification / Alternative check:Use ammeter placement thought-experiment: moving a series ammeter from one diode to the other cannot change the reading if there are no parallel branches.

Why Other Options Are Wrong:(b), (c), (d), and (e) suggest different currents despite a single series path, which violates the series current rule.

Common Pitfalls:Confusing different forward voltage drops with different currents in series; mixing series behavior with parallel branches where currents can differ.

Final Answer:500 mA