RBSE SS Voc Electrical and Elect common mistakes
RBSE SS Voc Electrical and Elect · Understand the traps and how to avoid them.
Calculating power with P = V × I when only resistance and current (or resistance and voltage) are given, then also misapplying the series/parallel rules by treating a parallel branch as if it carried the total current.
Candidates memorise P = VI as the only power formula and default to it even when current is unknown, and they carry the 'current is the same everywhere' intuition from series circuits into parallel networks where the supply current splits across branches.
What to do instead
Write the power wheel once at the top of the rough sheet: P = VI = I²R = V²/R. Before substituting, list the two quantities actually given, then pick the form that uses only those two. For networks, redraw and label branch currents separately, compute each branch with its own current, and sum branch powers to check against total P = V_supply × I_total.
Sizing a domestic circuit's protective device and conductor by adding appliance wattages and dividing by 230 V, ignoring diversity, then selecting a fuse or MCB rating above the cable's current-carrying capacity.
Candidates treat the protective device as protection for the appliance rather than for the cable, and they do not apply the concept that not all loads run simultaneously; the arithmetic feels complete once the total current is found.
What to do instead
State the rule explicitly in practice answers: the device protects the conductor, so its rating must not exceed the cable's tabulated current-carrying capacity, and the cable must carry the design current after diversity. Work a short example - total connected load, apply a diversity allowance, choose design current, then select cable from the table and a device at or just above design current but below cable capacity.
Describing earthing as 'connecting the body to earth so the current leaks away', without distinguishing the fault path that operates the protective device, and confusing protective earthing with equipotential bonding or with the neutral conductor.
Everyday language about 'leakage to earth' is absorbed from informal sources, so the candidate never forms the idea of a low-impedance fault loop that must carry enough current to trip the device within the required time.
What to do instead
Answer with the loop: fault from live conductor to exposed metalwork, via the protective conductor back to the source, producing fault current high enough to operate the overcurrent device or residual current device quickly, while limiting touch voltage on the metalwork. Separately define bonding as connecting extraneous conductive parts to the main earthing terminal to keep them at the same potential, and state that neutral is a current-carrying conductor, not a protective conductor.
Applying the transformer EMF equation or turns ratio with the wrong quantity - mixing rms and peak, using secondary current in the primary-side formula, or claiming a transformer changes frequency or changes the power rating.
Turns ratio is learned as a single fraction, so candidates apply it to whatever number appears, and the ideal-transformer assumption (power in equals power out) is not linked to the consequence that a step-up in voltage is a step-down in current.
What to do instead
Anchor on two invariants: frequency is unchanged, and V_p I_p = V_s I_s for an ideal transformer. Then write V_p/V_s = N_p/N_s = I_s/I_p and check the direction of the inequality - if the secondary voltage is higher, the secondary current must be lower. Specify which side each value belongs to in the working, and state that the EMF equation uses rms values of induced EMF.
Explaining why a three-phase induction motor is self-starting by saying 'the rotating field drags the rotor round', with no reference to slip, relative motion, or rotor current, and predicting that the rotor turns at synchronous speed at full load.
The rotating magnetic field is the memorable phrase, and the chain from relative speed to induced rotor EMF to rotor current to torque is a multi-step argument that candidates compress into one sentence under time pressure.
What to do instead
Write the causal chain in four labelled steps: stator currents produce a rotating field at synchronous speed; the stationary rotor sees a relative speed, so flux cuts the rotor bars and induces EMF and current; rotor current in the field produces torque; the rotor accelerates but can never reach synchronous speed because at zero slip there is no relative motion, no induced rotor EMF, and no torque. Add that slip s = (N_s − N_r)/N_s is small at full load but never zero.
Identifying a transistor's terminals or a diode's bias from a described circuit without checking polarity, and stating that a forward-biased silicon diode drops about 0.7 V while also claiming it blocks current in that direction.
Candidates rely on the arrow or on memorised 'P-N' letter order rather than tracing the applied potential, and the two facts (forward drop, forward conduction) are recalled separately without being reconciled into one behaviour.
What to do instead
For every device question, first mark which terminal is at the higher potential, then apply the rule: forward bias means the P side is positive relative to the N side, the diode conducts with roughly 0.7 V drop for silicon (about 0.3 V for germanium), and reverse bias blocks apart from small leakage. For a transistor, state the control relationship directly - base current controls collector current, emitter is common to both - before naming terminals.
Drawing or analysing a half-wave rectifier and describing the output as a smooth DC, or omitting the capacitor's role and the effect of load current on ripple, instead of stating the pulsating nature and the ripple frequency.
The word 'rectifier' suggests conversion to steady DC, and the smoothing capacitor is treated as an optional accessory rather than the component that determines ripple amplitude; ripple frequency is rarely linked to the supply frequency.
What to do instead
Describe the waveform explicitly: half-wave gives one output pulse per supply cycle, so ripple frequency equals supply frequency; full-wave and bridge give two pulses per cycle, so ripple frequency is twice supply frequency. Then add that the reservoir capacitor charges near the peak and discharges into the load between peaks, so ripple falls as capacitance and load resistance rise and rises as load current rises. Sketch the waveform with the ripple exaggerated.
Using a multimeter to measure voltage while the leads remain in the current sockets, or measuring resistance on a circuit that is still energised, and reading a value without checking the range or the AC/DC setting.
Candidates focus on the quantity they want and not on the meter's configuration; the current terminals are easy to leave in place, and resistance measurement on a live circuit seems harmless because the meter 'just reads'.
What to do instead
Adopt a fixed routine before every measurement: confirm function and range on the dial, confirm lead sockets match the function (voltage and resistance in the common and voltage sockets, current in the current socket), verify the circuit is isolated and capacitors discharged for resistance and continuity tests, and return leads to the voltage sockets immediately after any current measurement. State this sequence in answers about fault finding.
Describing soldering technique as 'apply the iron and the solder together' without mentioning tinning, tip contact area, or the difference between a good fillet and a cold joint, and using a high-wattage iron on heat-sensitive components.
The physical skill is learned by imitation, so the written description collapses to a single step, and joint quality is judged by 'it looks joined' rather than by wetting angle and surface appearance.
What to do instead
Write the sequence: clean and tin the tip, heat the joint (pad and lead) until both reach temperature, feed solder into the joint not onto the tip, remove solder then iron, and do not move the joint while it solidifies. Define a good joint as a shiny concave fillet with visible wetting, and a cold joint as dull, lumpy, or balled with poor wetting. Match iron power and tip size to the thermal mass, and mention heat shunts or low temperature for sensitive parts.
Answering a costing or self-employment question with only material cost, omitting labour, overheads, depreciation of tools, contingency and profit margin, and presenting a quotation without scope, terms, or a stated validity period.
Candidates equate 'cost' with what they paid for parts, because that is the only figure they have personally handled, and they have never seen a quotation document broken down, so they reproduce a single total.
What to do instead
Build every estimate in named lines: materials with wastage allowance, labour hours at a stated rate, overheads, tool depreciation or hire, transport, contingency, then profit as a percentage of total cost. Present the quotation with scope of work, exclusions, payment terms, validity period, and a note that variations are charged separately. Practise one full worked example so the structure is automatic.