Fully controlled bridge with RLE load – effect of reversing the battery EMF A fully controlled bridge converter feeds an R–L–E load. With one battery polarity, the required firing angle is 47°. If the battery connections are reversed (E changes sign), how will the required firing angle change to maintain operation?

Introduction / Context:
In an R–L–E load, E represents a back EMF or battery. The average DC output of a fully controlled bridge is Vdc = Vd0 * cos α (for continuous current), opposed or aided by E depending on its polarity. Reversing E alters the required α for the same current direction and magnitude.

Given Data / Assumptions:

• Line-commutated 1-phase bridge (concept extends similarly to 3-phase with different Vd0).
• Continuous current approximation.
• Initial α = 47° for given E.

Concept / Approach:

Converter must supply Vdc to overcome R*I plus the opposing EMF. If E reverses sign so that it opposes converter output more strongly (i.e., acts against power flow), the bridge must reduce its average output (more retarded firing) or even go toward inversion to keep current conditions, which means a larger α is needed.

Step-by-Step Solution:

Verification / Alternative check:

Limit case: if E is large and opposing, α may need to exceed 90° (inversion region) to hold current constraints.

Why Other Options Are Wrong:

• Equal or less than 47° would imply the converter output need not compensate the added opposition, which is incorrect.
• “Either” ignores the defined direction of opposition after reversal.

Common Pitfalls:

Forgetting that cos α decreases as α increases; mixing up aiding vs opposing EMF effects.

Final Answer:

more than 47°