Vidyut Sahayak (Pareshan) Contractual common mistakes
Vidyut Sahayak (Pareshan) Contractual · Understand the traps and how to avoid them.
In superposition, deactivating a voltage source by opening it and a current source by shorting it, then adding the partial results with the wrong sign.
The deactivation rules are easily reversed under exam pressure, and students memorise 'kill the source' without linking it to the ideal-source definitions: zero volts means a short, zero amperes means an open circuit. Dependent sources are also wrongly deactivated.
What to do instead
Write the rule on the rough sheet before solving: independent V-source → short, independent I-source → open, dependent sources stay active. Redraw the circuit after each deactivation and mark the assumed current direction on every branch so algebraic signs are added deliberately, not by memory.
Computing induction-motor slip, rotor frequency or rotor speed using line frequency and nameplate rpm without converting to synchronous speed in rps, or using f directly instead of s·f.
The formulas Ns = 120f/P, s = (Ns − N)/Ns and fr = s·f are known, but candidates mix rpm with rps, forget that Ns depends on pole pairs versus poles, and treat rotor frequency as constant at 50 Hz.
What to do instead
Standardise units before substitution: f in Hz, Ns in rpm using 120f/P, then s as a pure ratio. Compute fr = s·f only after s is found. Practise a table of P = 2, 4, 6, 8 at 50 Hz and check that slip at full load typically lies between 2% and 6%.
In three-phase fault or per-unit calculations, mixing line and phase quantities, omitting √3, or changing base kVA and base kV without converting impedance by the square of the voltage ratio.
The per-unit system removes √3 only when the base voltages are chosen line-to-line and base power is three-phase; candidates carry phase impedances into line-voltage formulas and apply base-change factors linearly.
What to do instead
Fix base quantities first: S_base three-phase, V_base line-to-line, Z_base = V_base²/S_base. Convert all impedances to per unit before solving, then convert fault current back using I_base = S_base/(√3·V_base). Check that a balanced three-phase fault current is E/Z1 in per unit, with no extra √3.
Using the full-wave controlled-rectifier average-voltage formula for a half-wave circuit, or dropping the 1/2π factor, and ignoring the extinction angle in discontinuous conduction.
The formulas (Vm/2π)(1+cos α) and (Vm/π)(1+cos α) look similar, and candidates memorise one expression without linking it to the conduction interval and the number of pulses per cycle.
What to do instead
Derive or recall the average as the integral of the output over one supply cycle: half-wave conducts for one pulse, full-wave for two. Write the pulse number beside every formula and verify at α = 0: half-wave gives Vm/π, full-wave gives 2Vm/π. State whether the load is resistive or highly inductive before choosing the formula.
Applying a current-transformer ratio in the wrong direction, or treating the secondary ammeter reading as the primary current, and ignoring CT burden when selecting the meter.
Ratio notation such as 100/5 means primary rated current to secondary rated current; students multiply when they should divide, and they forget that the secondary current depends on the primary current divided by the ratio, not on the burden.
What to do instead
Set up a ratio equation: I_primary/I_secondary = rated primary/rated secondary. With 100/5, multiply the ammeter reading by 20. Separately check burden: the connected VA must not exceed the CT's rated burden at the specified accuracy class, otherwise ratio and phase errors increase.
Classifying system type and computing steady-state error by applying the velocity-error formula to a step input, or forgetting that a type-0 system has finite step error but infinite ramp error.
The error constants Kp, Kv and Ka are memorised as a list, but candidates do not connect system type (number of integrators in the open loop) to which constant is finite, zero or infinite.
What to do instead
Count poles at the origin first, then build a small table: type 0 gives finite Kp and zero Kv; type 1 gives infinite Kp and finite Kv. Use the matching input: step with Kp, ramp with Kv, parabola with Ka. Check the limiting behaviour rather than substituting blindly.
Applying the virtual-short and virtual-ground assumptions to an op-amp used as a comparator or in positive feedback, and assuming zero input current in circuits without negative feedback.
The ideal op-amp rules V+ = V− and I+ = I− = 0 hold only in closed-loop negative feedback with finite output; candidates apply them to every op-amp symbol, including open-loop comparator and Schmitt-trigger circuits.
What to do instead
Before using virtual ground, confirm the feedback path returns to the inverting input and the amplifier is not saturated. For comparators, analyse the output as saturating at the supply rails and compare input voltages directly. For Schmitt triggers, identify positive feedback and derive the upper and lower threshold voltages separately.
Selecting a cable size or MCB rating from the connected load current alone, without applying derating for ambient temperature, grouping, enclosure, and without checking voltage drop and short-circuit withstand.
Nameplate current is the most visible number, and candidates treat the protective device as a simple current rating; they omit the installation method, the diversity factor, and the requirement that the protective device coordinate with cable current-carrying capacity.
What to do instead
Follow a fixed sequence: design current, then correction factors (ambient, grouping, thermal insulation), then choose the protective device so that Ib ≤ In ≤ Iz, then verify voltage drop against the permitted percentage and check the earth-fault loop impedance or short-circuit withstand. Show each factor explicitly in the answer.
Sizing a solar battery bank from daily energy divided by battery voltage only, ignoring depth of discharge, inverter efficiency and cable/charge losses, and confusing panel series voltage with parallel current.
Battery Ah capacity is not the same as usable Ah; candidates forget that lead-acid batteries should not be discharged below roughly 50% and that inverter and wiring losses add 15–25% to the required energy.
What to do instead
Use the full chain: daily energy demand ÷ inverter efficiency = DC energy; divide by depth of discharge to get installed capacity; divide by system voltage to get Ah. Apply a temperature and ageing margin. For the array, series strings set voltage and parallel strings add current; check that the string voltage stays within the MPPT window at high and low temperatures.
Spending disproportionate time on the Gujarati language and reasoning section, or attempting puzzles before the technical section, leaving insufficient time for calculation-heavy electrical questions.
Language questions feel quick and comfortable compared with numerical power-systems items, so candidates start there and lose the strategic advantage; puzzle-type reasoning can also absorb unlimited time without a stopping rule.
What to do instead
Adopt a fixed sectional plan: secure the easy language and general-awareness marks within a set time cap, then move to technical sections where accuracy is high. For reasoning, set a per-question limit; if the pattern is not visible within that limit, mark and skip. Reserve the final minutes for review of numerical answers and unit checks.