--- title: "DFRT and equivalent circuit of single phase IM" subject: "Electric Machines II" chapter: 3 key: "semester_5/electric_machines_ii:dfrt-equivalent-circuit" source: https://www.bctengineer.com/q/semester_5/electric_machines_ii/dfrt-equivalent-circuit site: BCT Engineering --- # DFRT and equivalent circuit of single phase IM Electric Machines II · Chapter 3: Fractional Kilowatt Motors - Asked in: 2072 Kartik, 8 marks - Video: [double revolving field theory of single phase induction motor | double revolving theory](https://www.youtube.com/watch?v=pXmPCSc9di0) — Mind of Engineer - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/electric_machines_ii/dfrt-equivalent-circuit - Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machines_ii/chapter-3 ## Question What is double field revolving theory in single phase induction motor? Explain the operation of single phase induction motor through its equivalent circuit. ## Short answer Double field revolving theory resolves a pulsating field into two counter-rotating fields of half amplitude. In the equivalent circuit, the rotor is represented by forward (slip s) and backward (slip 2-s) branches; net torque is I1^2(R_F-R_B)/omega_s. At standstill R_F=R_B, torque zero; at run the forward branch dominates, producing net torque. ## Answer **Double field revolving theory (DFRT)** states that a pulsating magnetic field of amplitude $\phi_m$ can be resolved into **two rotating fields**, each of amplitude $\phi_m/2$, rotating at synchronous speed $N_s$ in **opposite directions**. $$ \phi_m\cos\omega t \cos\theta = \frac{\phi_m}{2}\cos(\theta - \omega t) + \frac{\phi_m}{2}\cos(\theta + \omega t) $$ #### Torque from the two fields - If the rotor runs at speed $N$ in the forward direction, slip w.r.t. forward field: $s_f = s = \dfrac{N_s - N}{N_s}$. - Slip w.r.t. backward field: $s_b = \dfrac{N_s + N}{N_s} = 2 - s$. - Forward field gives torque $T_f$, backward field gives $T_b$ (opposite). Net torque $T = T_f - T_b$. - At standstill $s = 1$, $s_b = 1$: $T_f = T_b$, net torque zero, so the motor is **not self-starting**. Once rotating, $T_f > T_b$ and it runs on. ```text T | Tf | /\ Tnet = Tf - Tb | / \ ---+---------------------> N | \ / | \/ Tb ``` #### Equivalent circuit Since each rotating field is half the total, the rotor can be represented by **two half-rotors**: one acted on by the forward field (slip $s$) and one by the backward field (slip $2-s$). Magnetising reactance and rotor impedance are each split into halves. ```text I1 R1 X1 o--/\/\--mmm--+----------+ | 0.5Xm | 0.5R2'/s +--mmm--+--+--/\/\--mmm-- (Forward) V | | 0.5X2' +----------+ | 0.5Xm | 0.5R2'/(2-s) +--mmm--+--+--/\/\--mmm-- (Backward) | | 0.5X2' o-------------+----------+ ``` Forward and backward impedances: $$ \begin{aligned} Z_F &= R_F + jX_F = \frac{(j0.5X_m)\left(\frac{0.5R_2'}{s} + j0.5X_2'\right)}{\frac{0.5R_2'}{s} + j0.5(X_m + X_2')} \\ Z_B &= R_B + jX_B = \frac{(j0.5X_m)\left(\frac{0.5R_2'}{2-s} + j0.5X_2'\right)}{\frac{0.5R_2'}{2-s} + j0.5(X_m + X_2')} \end{aligned} $$ #### Operation through the equivalent circuit 1. Input current: $I_1 = \dfrac{V}{R_1 + jX_1 + Z_F + Z_B}$. 2. Air-gap powers: $P_{gF} = I_1^2 R_F$, $P_{gB} = I_1^2 R_B$. 3. Torques: $T_f = \dfrac{P_{gF}}{\omega_s}$, $T_b = \dfrac{P_{gB}}{\omega_s}$, net $T = \dfrac{P_{gF} - P_{gB}}{\omega_s}$. 4. Mechanical power developed: $P_m = (1-s)(P_{gF} - P_{gB})$. 5. Rotor copper loss: $s P_{gF} + (2-s) P_{gB}$. **Interpretation:** - At $s = 1$: $Z_F = Z_B$, so $P_{gF} = P_{gB}$ and $T = 0$ (no starting torque). - At normal slip (small $s$): $0.5R_2'/s$ is large, so $Z_F \gg Z_B$. Most of the voltage appears across the forward branch, the forward field is strong and the backward field is weak, giving positive net torque. - The backward field adds extra rotor copper loss and a double-frequency torque pulsation, so a single-phase motor has lower efficiency and is noisier than a three-phase motor. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. If the forward slip is s, the slip with respect to the backward field is: A. s B. 1 - s C. 2 - s D. 1 + s Answer: C. s_b = (N_s + N)/N_s = 2 - s. (Key point: Backward slip) 2. How is the rotor represented in the single-phase equivalent circuit? A. Single full branch at slip s B. Two half-rotors, one at slip s and one at slip 2 - s C. Only the forward half D. Two full rotors in series with the supply Answer: B. Each rotating field is half the total, so half values of X_m and rotor impedance are used for each field. (Key point: Equivalent circuit) 3. Net torque from the equivalent circuit is: A. (P_gF + P_gB)/omega_s B. (P_gF - P_gB)/omega_s C. P_gF x P_gB D. P_gB/omega_s Answer: B. T = T_f - T_b = (P_gF - P_gB)/omega_s. (Key point: Net torque) 4. At normal slip Z_F is much larger than Z_B because: A. 0.5R2'/s is large, so most voltage drops across the forward branch B. 0.5R2'/(2-s) is large C. X_m is zero D. The backward field is stronger Answer: A. A large forward branch resistance gives a strong forward field and a weak backward field. 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