Electricity is the movement of electric charge — in most circuits, this means electrons drifting through a conducting wire. Think of it like water flowing through a pipe: voltage (V) is the pressure pushing the water, current (I) is how fast the water flows (litres per second), and resistance (R) is the narrowness of the pipe that opposes the flow.
This analogy, which you would actually use with a Class 7 student, maps directly to Ohm's Law:
V = I × R
A circuit is simply a closed loop that allows charge to flow from a source (battery), through components (bulbs, resistors, switches), and back. Break the loop anywhere and current stops — that is the principle behind a switch, and it is also the first thing you demonstrate to Class 6 students when they build a simple torch circuit.
Two fundamental configurations matter for the exam:
In real life, your home wiring is parallel — so that switching off one light does not kill all the others. A string of old-style festive lights was series — when one bulb failed, the whole string went dark. That contrast is a standard Class 7 classroom discussion point, and it shows up in CTET questions disguised as numerical problems.
Materials also matter. Conductors (copper, aluminum) have low resistance and allow easy current flow. Insulators (rubber, plastic) have very high resistance and block current. Semiconductors (silicon, germanium) sit in between — their conductivity can be precisely controlled, which is why every transistor and microchip is made from them. At the Class 6-8 level, knowing the classification and one clear example of each is enough.
Ohm's Law — V = IR — holds for ohmic conductors (metallic wires at constant temperature). It does not hold for semiconductors, electrolytes, or filament bulbs when they are heating up (because resistance changes with temperature). CTET occasionally tests whether a teacher knows this boundary condition.
In a series circuit with n resistors:
R_total = R₁ + R₂ + R₃ + ... + Rₙ
The same current I flows through all. Voltage divides in proportion to resistance:
V₁/V₂ = R₁/R₂
This voltage-divider relationship is the fastest way to solve questions like PYQ 6a2c204a8019520d6645dbf6 — you do not need to find total voltage at all.
In a parallel circuit:
1/R_total = 1/R₁ + 1/R₂ + 1/R₃ + ...
The equivalent resistance is always less than the smallest individual resistance — this is your instant sanity-check on any parallel answer. Voltage across every branch is identical. Current in each branch: I_branch = V/R_branch.
For exactly two resistors in parallel, the product-over-sum shortcut applies:
R_parallel = (R₁ × R₂) / (R₁ + R₂)
The strategy is always: collapse the parallel section first, then treat the result as a series resistor. This two-step reduction is the template for every mixed-circuit PYQ.
Step 1 — Find R_p for the parallel part.
Step 2 — Add R_p to any series resistors to get R_total.
Step 3 — Apply I = V / R_total.
R = ρL/A
where ρ (rho) is the resistivity of the material, L is length, and A is cross-sectional area.
The wire-stretching problem (PYQ 6a2c204a8019520d6645dbf3) is a classic trap. When a wire is stretched to double its length, volume is conserved (Volume = L × A). So if L doubles, A must halve. Substituting:
R_new = ρ(2L)/(A/2) = 4 × ρL/A = 4R_original
This means resistance becomes 4 times, not 2 times — a very common mistake at every level.
When two identical cells are in series aiding (+ to − connection), net EMF = sum of individual EMFs.
When in series opposition (+ to + or − to −), their EMFs cancel:
EMF_net = EMF₁ − EMF₂
For two identical cells in opposition, EMF_net = 0. No current flows. This is conceptually important — the cells are still present, but they neutralise each other.
Three equivalent expressions for power — know all three:
P = VI = I²R = V²/R
In a parallel circuit, every branch has the same voltage V, so the most convenient formula is P = V²/R per component. Total power is the sum. In a series circuit, the same current I flows, so P = I²R per component is more convenient.
| Category | Resistivity Range | Examples |
|---|---|---|
| Conductor | Very low (10⁻⁸ to 10⁻⁶ Ω·m) | Copper, silver, aluminium, graphite |
| Semiconductor | Intermediate (10⁻⁴ to 10⁰ Ω·m) | Silicon, germanium |
| Insulator | Very high (10⁶ to 10¹⁶ Ω·m) | Rubber, glass, plastic, dry wood |
Note that graphite is a conductor (used in electrodes and pencils), not an insulator — a common distractor. Silicon is the canonical semiconductor answer for any CTET item about electronic devices.
500 mA = 0.5 A (not 0.002 A — see the explanation in PYQ 6a2c204a8019520d6645dbf9)1 kΩ = 1000 Ω1 mW = 0.001 Wmilli-ampere conversions are the single most common arithmetic trap in electricity PYQs.
Before you calculate, note the smallest resistor in a parallel group. Your answer must be less than that value. If you get a parallel resistance of 4 Ω when the individual resistors are 4 Ω, 6 Ω, and 12 Ω — you know immediately you made an error. This single rule eliminates the most common parallel-circuit mistake with zero calculation. Standard method: finish the calculation, then check. This rule: pre-checks the answer in 2 seconds.
Any time a wire is stretched to n times its original length (volume conserved), resistance becomes n² times. Length doubles → R × 4. Length triples → R × 9. This is because both length increases (×n) and area decreases (÷n), multiplying together to give ×n². Apply directly: standard method requires writing ρ(nL)/(A/n) = n²(ρL/A) — 4 steps. Pattern: state "doubled length → ×4" in 1 step.
In a series circuit, voltage splits in the same ratio as resistances. If R₁:R₂ = 5:10 = 1:2, and you know V₁ = 4 V, then V₂ = 8 V immediately — no need to find total voltage or current first. Standard method: find I, then V₂ = IR₂ — 3 steps. Ratio method: spot 1:2, scale V₁ — 1 step. Saves roughly 20 seconds per question.
In a parallel circuit with a known supply voltage, calculate each branch's power as P = V²/R and sum them. Do not find equivalent resistance, then find current, then find power — that is three unnecessary steps. For V = 12 V and R = 4 Ω: P = 144/4 = 36 W. Done. Standard method (find R_eq → I_total → P = V×I): 4 steps. Direct method: 1 step per branch.
When two identical cells face each other (positive terminal to positive terminal, or negative to negative), write the answer as 0 V immediately. There is nothing to calculate. The options will include "3 V" (series aiding), "1.5 V" (one cell alone), and "0 V". Eliminate 3 V and 1.5 V in 3 seconds — pick 0 V. This works whenever the question says "series opposition" with identical cells.
When you see an electricity numerical in the exam hall, run this decision tree:
Step 1 — Identify circuit type.
Is it purely series? Purely parallel? Mixed?
Step 2 — Mixed circuits: collapse parallel section first.
Use 1/R_p = 1/R₁ + 1/R₂ + ... or product-over-sum for two resistors. Then add series resistors.
Step 3 — Apply I = V/R_total.
This gives the total (main line) current.
Step 4 — What is the question actually asking?
V = IR_element.P = V²/R for parallel branches, P = I²R for series elements.Step 5 — Unit check.
Is current in mA? Convert before plugging into V = IR. Is resistance in kΩ? Convert to Ω.
Step 6 — Sanity check.
Parallel R must be less than smallest R. Series R must be greater than largest R. Power must be positive.
This six-step flow handles every electricity numerical at the CTET level in under 90 seconds.
Why this question: Tests the mixed series-parallel circuit template — the most common circuit configuration in CTET numericals.
Solving path: Collapse the 6 Ω and 3 Ω parallel pair first: 1/R_p = 1/6 + 1/3 = 3/6, so R_p = 2 Ω. Total resistance = 2 + 2 = 4 Ω. Current = 12/4 = 3 A. Notice that the parallel pair and the series resistor happen to both equal 2 Ω — that is a deliberate design that makes the arithmetic clean but can mislead you into thinking the answer is 2 A if you stop one step early.
Why this question: The wire-stretching problem is a perennial trap. Almost every aspirant instinctively says "24 Ω" by doubling — the correct answer is 4 times, not 2 times.
Solving path: Volume conserved: L × A = constant. If L → 2L, then A → A/2. New R = ρ(2L)/(A/2) = 4ρL/A = 4 × 12 = 48 Ω. The answer is 48 Ω, not 24 Ω. The distractor 24 Ω is there specifically to catch people who only account for the length doubling.
Why this question: Tests power calculation in a parallel circuit and the choice between P = VI and P = V²/R. The direct-per-branch approach is faster than finding equivalent resistance.
Solving path: Each bulb sees the full 12 V. P₁ = 144/4 = 36 W, P₂ = 144/6 = 24 W, P₃ = 144/12 = 12 W. Total = 72 W. Cross-check: equivalent parallel R = 2 Ω (same set of resistors as PYQ 6a2c204a8019520d6645dbf8), so total current = 12/2 = 6 A, total power = 12 × 6 = 72 W. Both routes give the same answer — use whichever is faster for you.
Why this question: Tests conceptual understanding of cell polarity — a topic that appears in Class 8 science and is often set as a "trick" item in CTET.
Solving path: Series opposition means the cells fight each other. Net EMF = 1.5 − 1.5 = 0 V. No current flows. Eliminate 3 V (that is series aiding) and 1.5 V (single cell). Answer is 0 V. This requires zero arithmetic — it is entirely conceptual.
Why this question: Tests the unit-conversion trap (mA to A) that the spec explicitly flags as a common error.
Solving path: Convert 500 mA = 0.5 A. Apply Ohm's Law: R = V/I = 2.5/0.5 = 5 Ω. The distractor 1250 Ω is what you get if you mistakenly treat 500 mA as 0.002 A (confusing milli with micro). The distractor 2.5 Ω looks plausible because V = 2.5 V. Always write the unit conversion as a explicit step.
Wire stretching — doubling length gives ×2 resistance, not ×4. This is the single most-missed wire problem. Remember: area halves when length doubles (volume conservation), so both changes multiply to give ×4.
Converting mA to A incorrectly. 500 mA = 0.5 A, not 0.002 A. If you write 0.002 A, you are confusing milliampere with microampere. Write the conversion explicitly every single time.
Parallel resistance greater than the smallest resistor. If your calculated parallel equivalent is bigger than any individual resistor, you have made an arithmetic error. This is impossible — equivalent parallel resistance is always smaller.
Applying P = I²R to parallel circuits without finding branch current. The shortcut is P = V²/R using the common branch voltage. Using P = I²R with total current in a parallel circuit gives a completely wrong answer.
Series opposition mistaken for series aiding. "Series" alone means aiding (EMFs add). "Series opposition" is the special case where EMFs cancel. Read the connection description carefully — look for "positive to positive" or "negative to negative" as the signal.
Graphite classified as insulator. Graphite is a conductor (used in electrodes, pencils conduct electricity). Rubber and plastic are insulators. Graphite-as-insulator is a standard wrong-answer option in material classification questions.