What is the formula of the ratio of the lower limit to the upper limit of current?

Question

Examoogle · Accepted Answer

Option D: All of these

Explanation & Concepts:
In DC motor starters, particularly three-point and four-point starters with graded resistance sections, the upper limit of current $I_1$ and lower limit of current $I_2$ are held between constant limits during the starting process. The ratio of the lower limit to upper limit of current, $\frac{I_2}{I_1}$, is related to the resistance steps by $\frac{I_2}{I_1} = \frac{R_2}{R_1} = \frac{R_3}{R_2} = \dots = \frac{R_{n+1}}{R_n} = \left(\frac{R_{n+1}}{R_1}\right)^{\frac{1}{n}}$. Since armature resistance $r_a = R_{n+1}$, option B is an equivalent representation, making 'All of these' the correct answer.

Working Principle: During starting, as the starter handle is moved from stud to stud, the current fluctuates between $I_1$ (upper limit) and $I_2$ (lower limit). To keep current fluctuations uniform across all studs, the resistance of the sections forms a geometric progression. Taking the product of $n$ equal resistance ratio steps yields $\left(\frac{I_2}{I_1}\right)^n = \frac{R_{n+1}}{R_1}$, which gives $\frac{I_2}{I_1} = \left(\frac{R_{n+1}}{R_1}\right)^{\frac{1}{n}} = \left(\frac{r_a}{R_1}\right)^{\frac{1}{n}}$.

Key Formulas: $\frac{I_2}{I_1} = \frac{R_{k+1}}{R_k}$ — Ratio of lower to upper current limit for each stud section; $\frac{I_2}{I_1} = \left(\frac{R_{n+1}}{R_1}\right)^{\frac{1}{n}} = \left(\frac{r_a}{R_1}\right)^{\frac{1}{n}}$ — Current ratio expressed in terms of total starting resistance and armature resistance

Key Points: To minimize current surges, starter resistance sections are designed such that resistance values form a geometric progression.; The total number of resistance steps is denoted by $n$, with $n+1$ total resistance values from $R_1$ (total resistance) down to $R_{n+1} = r_a$ (armature resistance).; Since $\frac{R_2}{R_1} = \frac{R_3}{R_2} = \dots = \frac{R_{n+1}}{R_n}$, the product of these $n$ ratios is $\frac{R_{n+1}}{R_1} = \left(\frac{I_2}{I_1}\right)^n$.

Advantages: Ensures uniform current limits during motor acceleration; Prevents thermal damage to armature windings by limiting peak currents

Disadvantages: Requires precise calculation and tapering of resistance values; Energy is dissipated as heat in starting resistors during start-up

Applications: Design of three-point and four-point starters for DC shunt and compound motors; Design of rotor resistance starters for 3-phase slip ring induction motors

Additional Info: Option A gives the step-by-step resistance ratio equivalence to current ratio.. Option B substitutes armature resistance $r_a$ in place of $R_{n+1}$.. Option C presents the standard nth root form involving total resistance $R_1$ and final resistance step $R_{n+1}$.. Therefore, options A, B, and C are all valid expressions for the current limit ratio.

Answer: D is correct — All three equations correctly represent the ratio of lower current limit to upper current limit ($\frac{I_2}{I_1}$) in DC motor starter design.

Exam Tip: In competitive exams like GATE and ESE, remember that if starter resistance steps form a GP, the common ratio $K = \frac{I_2}{I_1} = \left(\frac{r_a}{R_1}\right)^{1/n}$. Number of studs is always $n+1$ where $n$ is the number of resistance sections.

Examoogle · Answer

Option A: I2I1=R2R1=R3R2=R4R3=Rn+1Rn

Examoogle · Answer

Option B: I2I1=raR11n

Examoogle · Answer

Option C: I2I1=Rn+1R11n

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