Basic Electrical and Electronic Fundamentals
What this domain tests
The exam checks numerical fluency with Ohm's law and power, series and parallel circuit behaviour, and safe recognition of AC and DC. Questions often give a supply and two or three resistors and ask for current, voltage drop or power.
Rules to apply
- Ohm's law: V = IR; power: P = VI = I²R = V²/R.
- Series: R_total = R1 + R2 + ...; the same current flows through every element; voltages add.
- Parallel: 1/R_total = 1/R1 + 1/R2 + ...; voltage is the same across branches; branch currents add.
- A voltage divider gives V_out = V_supply × R2/(R1+R2) only when the output is unloaded.
- Domestic AC in India is 230 V, 50 Hz; a battery is DC. The 230 V value is RMS, so the peak is about 325 V.
- The earth wire is a safety conductor, not a normal current-carrying return.
Worked example
A 12 V supply feeds 100 Ω and 200 Ω in series. R_total = 300 Ω, so I = 12/300 = 0.04 A = 40 mA. Drop across 100 Ω = 0.04 × 100 = 4 V; across 200 Ω = 8 V. Total power = 12 × 0.04 = 0.48 W. Check: 4 V + 8 V = 12 V. If the same resistors were in parallel, R_total = (100 × 200)/(300) ≈ 66.7 Ω and the supply current would be about 0.18 A.
Traps
- Entering milliamps into V = IR without converting to amps.
- Adding parallel resistors directly instead of using reciprocals.
- Forgetting that a loaded divider no longer follows the simple ratio.
- Calling 230 V the peak value.