A fully controlled single-phase bridge converter feeds a continuous-current RLE load (E is back-emf, R is resistance, L large). The source is v(t) = V_m sin(ωt). Derive the firing-angle relation and choose the correct expression for cosα.

Introduction / Context:Single-phase fully controlled rectifiers feeding DC motor-type loads (RLE) are classic in drives. For continuous current (large L), the average output voltage relates directly to firing angle α, allowing us to solve for α given load back-emf and drop across R.Given Data / Assumptions:

• v_s(t) = V_m sin(ωt), single-phase.
• Fully controlled bridge (four-quadrant gating within rectifier).
• Continuous current (inductor large), so output ripple is small.
• Average DC output V_dc = E + I_0 R.

Concept / Approach:For a fully controlled bridge with sinusoidal source and continuous current: V_dc = (2 V_m / π) * cosα. This comes from integrating the sinusoid over conduction interval with firing delay α. Then set V_dc equal to the DC load requirement E + I_0 R and isolate cosα.Step-by-Step Solution:1) Use the known average: V_dc = (2 V_m / π) * cosα.2) KVL on DC side: V_dc = E + I_0 R (inductor drop averages to zero in steady state).3) Equate and solve: (2 V_m / π) * cosα = E + I_0 R.4) Hence cosα = (π / (2 V_m)) * (E + I_0 R).Verification / Alternative check:Dimensional consistency: right-hand side is dimensionless since E and I_0 R are volts and the multiplier is π/(2 V_m) with units 1/V.Why Other Options Are Wrong:

• π/V_m factor: off by a factor of 2.
• Replacing + with −: violates KVL unless direction conventions differ (here continuous motoring mode uses +).
• Division by V_m alone: missing π/2 scaling that arises from averaging.

Common Pitfalls:

• Forgetting that average inductor voltage is zero in steady state.
• Mixing RMS and peak values; formula requires V_m (peak).

Final Answer:cosα = (π / (2 V_m)) * (E + I_0 R).