In a bipolar junction transistor (BJT), does the collector-to-emitter voltage satisfy VCE = VBE + VCB by Kirchhoff’s voltage law around the three-terminal loop? Choose the most accurate statement.

Introduction / Context:
Kirchhoff’s voltage law (KVL) applies to transistor junctions just as it does to passive components. Knowing the algebraic relationships between VBE, VCB, and VCE is useful for quick checks and for translating measurements between different terminal pairs.

Given Data / Assumptions:

• BJT in any operating region where terminal voltages are well-defined.
• Voltages referenced consistently (polarities defined from first terminal to second).
• No need to know current gain beta to write KVL identities.

Concept / Approach:
The three terminal voltages relate by loop addition around the C–B–E loop. Tracing from collector to emitter through the base node gives VCE = VCB + VBE. This identity is purely algebraic and holds regardless of region of operation (cutoff, active, saturation), provided sign conventions are consistent.

Step-by-Step Solution:

Verification / Alternative check:
Check with typical values in forward-active operation: VBE ≈ 0.7 V, VCB may be a few volts reverse (positive if defined C→B). Summing gives VCE of a few volts, consistent with practical biasing.

Why Other Options Are Wrong:

• Option b changes a sign without justification; it contradicts KVL as written with standard polarities.
• Option c mixes reversed polarity definitions (VBC, VEB) leading to confusion.
• Option d incorrectly implies dependence on beta, which affects currents, not these voltage identities.

Common Pitfalls:
Inconsistent polarity definitions (e.g., using VBC instead of VCB without flipping the sign) or mixing measured and defined polarities.

Final Answer:
Correct — VCE = VBE + VCB.