Current electricity in NET tests Ohm's law, resistivity, series and parallel combinations, electrical power, and emf with internal resistance. The questions are short circuits you can solve in under a minute if you know how R scales with length and area, and which power formula to pick.
Key facts
- Physics is 30% of NET-Engineering, a derived 60 of the 200 MCQs.
- Current electricity shares a first-year chapter with electrostatics. Our syllabus data estimates about 8 questions from that combined chapter.
- Syllabus: Ohm's law, resistivity, series and parallel networks, and electrical power.
Source: NUST NET-Engineering composition and the SailToNUST physics syllabus, checked October 2026
What the chapter covers
The first-year chapter "Electrostatics and Current Electricity" starts with charges and capacitors, then moves to current. This post covers the current half: drift of charge, Ohm's law, resistivity and its change with temperature, combinations of resistors, emf and internal resistance, power, and the Wheatstone bridge. Coulomb's law and capacitors are in the electrostatics guide.
Formulas you need
Current I = Q/t A = C s⁻¹
Drift I = nAvq
Ohm's law V = IR Ω = V A⁻¹
Resistance of a wire R = ρL/A ρ in Ω m
Temperature R_T = R₀(1 + αΔT) α in K⁻¹
Conductance G = 1/R siemens (S)
Series R = R₁ + R₂ + ... same current
Parallel 1/R = 1/R₁ + 1/R₂ + ... same voltage
Two in parallel R = R₁R₂ / (R₁ + R₂)
n equal resistors R series nR, parallel R/n
Power P = VI = I²R = V²/R W
Energy E = Pt; 1 kWh = 3.6 × 10⁶ J
emf and internal resistance
I = ε / (R + r)
Terminal voltage V = ε − Ir
Maximum power to load when R = r
Wheatstone bridge balance P/Q = R/S
Worked MCQs
1. Two wires in parallel. Wires P and Q are the same material and length. P has diameter 2 mm and Q has diameter 1 mm. Both connect in parallel across the same supply. What is the ratio of current in P to current in Q?
(a) 1/4 (b) 4 (c) 2 (d) 1/2
R = ρL/A and A ∝ d²
d_P = 2 d_Q so A_P = 4 A_Q so R_P = R_Q / 4
Same V across both: I ∝ 1/R
I_P / I_Q = 4
Answer (b). People who scale with d instead of d² pick (c).
2. Stretching a wire. A wire of resistance R is stretched to twice its length. Its volume stays the same. What is the new resistance?
(a) R/2 (b) 2R (c) 4R (d) R
Volume = AL constant
L → 2L means A → A/2
R = ρL/A → ρ(2L)/(A/2) = 4R
Answer (c). The general rule: stretch by a factor n and resistance goes up n². Picking (b) means you forgot the area shrinks.
3. Bulb on half voltage. A bulb is rated 100 W, 220 V. What power does it draw on a 110 V supply? Assume its resistance does not change.
(a) 50 W (b) 25 W (c) 100 W (d) 200 W
R = V²/P = 220² / 100 = 484 Ω
P = 110² / 484 = 25 W
Answer (b). Halve the voltage and power drops to a quarter, because P = V²/R with R fixed.
4. Battery with internal resistance. A cell of emf 12 V and internal resistance 0.5 Ω drives current through a 5.5 Ω resistor. What is the terminal voltage?
(a) 12 V (b) 11 V (c) 10 V (d) 6 V
I = ε / (R + r) = 12 / 6 = 2 A
V = ε − Ir = 12 − 2 × 0.5 = 11 V
Answer (b). A voltmeter across the cell reads 11 V, not 12 V, while current flows. It only reads the full emf on open circuit.
5. Smallest resistance from two resistors. What is the smallest resistance you can make from a 6 Ω and a 12 Ω resistor?
(a) 2 Ω (b) 4 Ω (c) 6 Ω (d) 18 Ω
Parallel: (6 × 12) / (6 + 12) = 72 / 18 = 4 Ω
Answer (b). A parallel combination is always smaller than the smallest single resistor in it. If you got an answer above 6 Ω for a parallel pair, check your arithmetic.
Traps that cost marks
- Diameter versus radius. Resistance goes as 1/d² and 1/r² alike, so ratios come out the same. But if a question gives a diameter and you need an area in m², halve it first.
- Bulbs in series. Two bulbs rated 40 W and 100 W at 220 V, joined in series: the 40 W bulb glows brighter. It has the higher resistance, and in series the same current flows, so P = I²R favours the bigger R.
- Resistivity is a material property. Cutting a wire in half halves its resistance but leaves ρ unchanged. Only temperature and material change ρ.
- Metals versus semiconductors. Metal resistance rises with temperature (α positive). Semiconductor and carbon resistance falls (α negative).
- Ideal meters. An ideal ammeter has zero resistance and goes in series. An ideal voltmeter has infinite resistance and goes in parallel. Swap them and the circuit question is designed to catch you.
- kWh is energy, not power. A 1.5 kW heater run 4 hours a day for 30 days uses 1.5 × 4 × 30 = 180 kWh.
Speed tips for the exam
Learn the scaling rules instead of plugging numbers. R ∝ L, R ∝ 1/d², stretched by n gives n², P ∝ V² at fixed R, P ∝ I² at fixed R. Most NET current-electricity questions are ratio questions, and ratios take ten seconds.
For networks, redraw the circuit with junctions labelled before you combine anything. Two resistors are in parallel only if both ends connect to the same two junctions.
For equal resistors in parallel, divide by the count: three 9 Ω resistors in parallel give 3 Ω. Mixed values take product over sum, two at a time.
Electricity questions from this chapter and from electromagnetism often sit next to each other in a paper. The physics formula sheet puts both on one page.
FAQ
Is Kirchhoff's law in NET?
The syllabus summary lists series and parallel networks. Kirchhoff's rules are how you solve networks that are neither, so know the junction and loop rules and be able to use them on a two-loop circuit.
What happens to resistance when a wire is stretched?
If the wire is stretched n times its length at constant volume, its resistance becomes n² times the original. Doubling the length gives 4R.
Why does a battery's voltage drop when it supplies current?
Some of the emf is used pushing current through the cell's own internal resistance. The terminal voltage is ε − Ir, which falls as current rises.
Is current electricity easy in NET?
It is one of the most predictable chapters. The same five or six setups repeat, so timed practice pays off faster here than in most chapters.
