Single-phase fully controlled bridge feeding an R–L–E load (continuous current) Given v_s(t) = Vm sin(ωt), firing angle α, and average current I0, which relation is correct?

Introduction / Context:For a fully controlled bridge supplying an R–L–E load (e.g., a DC motor back-emf), the average DC output voltage depends on firing angle α and equals the sum of the average resistive drop and the back-emf. This relation is the cornerstone for speed and torque control in drives.

Given Data / Assumptions:

• Input: v_s(t) = Vm sin(ωt).
• Fully controlled bridge (four SCRs), continuous current.
• Ideal devices, no overlap; average load current = I0; series R and back-emf E.

Concept / Approach:

For continuous current and no overlap, the average output voltage is V_avg = (2Vm/π) cos α. KVL for the DC side gives V_avg = R I0 + E. Combining these two expressions yields the desired relation for cos α in terms of R, I0, E, and Vm.

Step-by-Step Solution:

Verification / Alternative check:

Limit checks: with E = 0 and α = 0°, cos α = 1 ⇒ V_avg = 2Vm/π, the well-known maximum. Increasing α reduces V_avg linearly with cos α as expected.

Why Other Options Are Wrong:

Options with incorrect proportionality (e.g., multiplied by 2Vm/π instead of divided) or sign errors (E − R I0) do not satisfy both the converter equation and DC KVL simultaneously.

Common Pitfalls:

Mixing up Vm (peak) with V_rms; forgetting continuous conduction assumption; ignoring the back-emf term.

Final Answer:

cos α = (π/(2Vm)) * (R I0 + E)