A wire of resistance R is stretched uniformly to twice its original length. Its new resistance is:

R, since the material has not changed

Two bulbs rated 100 W and 60 W for the same mains voltage are connected in parallel to that supply. Which glows brighter, and why?

The 100 W bulb, because at the same voltage P = V²/R and it has the lower resistance

The 60 W bulb, because it has the higher resistance

The 100 W bulb, because at the same voltage P = V²/R and it has the lower resistance

Both equally, because they share the same voltage

The 60 W bulb, because it draws the larger current

A potentiometer is preferred to a voltmeter for measuring the emf of a cell because:

At balance it draws no current from the cell, so no internal drop occurs

It has a lower resistance than a voltmeter

At balance it draws no current from the cell, so no internal drop occurs

It can measure alternating as well as direct voltages

Its scale is longer, so the reading is easier to see

Why does a bulb light almost instantly when the switch is closed, even though the drift velocity of electrons is only about a millimetre per second?

Because the electric field is set up along the whole circuit almost at the speed of light, and every electron starts drifting at once

Because the electrons in the filament travel at the speed of light

Because the electric field is set up along the whole circuit almost at the speed of light, and every electron starts drifting at once

Because the current in the filament is very large

Because the filament already stores electrons before the switch is closed

Current Electricity — Resistance, Cells, Networks and the Bridge

Current electricity is the most reliably scoring block in this paper because its questions are almost all substitutions. Four of them recur: reduce a series-parallel network, use V = IR or P = V²/R once, find terminal voltage from internal resistance, or apply the balance condition of a Wheatstone bridge or a potentiometer. What CUET does not do is set a network needing three simultaneous equations — that is the JEE register, and the JEE chapter in this course is where it lives. This chapter is written to the substitutions, with the conditions attached: resistivity is a property of the material and resistance is a property of the piece, and drift velocity is tiny while the signal is not.

Resistance, Resistivity and What Changes With What

R = ρL/A, and the distinction the paper tests is that ρ belongs to the material while R belongs to the piece: stretch a wire and ρ is unchanged, but L rises and A falls, so R rises. Drift velocity is v_d = I/(nAe) and is of the order of a millimetre per second — the standard question asks why a bulb lights instantly if the electrons crawl, and the answer is that the electric field is established through the circuit at nearly the speed of light while the electrons themselves drift slowly. Resistivity of a conductor rises with temperature; for a semiconductor it falls, which is the one-line contrast CUET likes.

The substitutions the paper actually asks for
Asked as	Use	The catch
Series combination	`R = R₁ + R₂ + …`; the CURRENT is common	The total is larger than the largest resistor. Voltage divides in the ratio of the resistances.
Parallel combination	`1/R = 1/R₁ + 1/R₂ + …`; the VOLTAGE is common	The total is smaller than the smallest resistor. Two equal resistors in parallel give half of one.
Power in a resistor	`P = VI = I²R = V²/R`	At constant V use V²/R, so the SMALLER resistor dissipates more; at constant I use I²R, so the larger one does. Read which is fixed.
Cell with internal resistance	`I = ε/(R + r)`, terminal `V = ε − Ir`	Terminal voltage equals the emf only on open circuit, and exceeds it only while the cell is being charged.
Wheatstone bridge	Balanced when `P/Q = R/S`	At balance the galvanometer reads zero and its own resistance stops mattering — that is why the method is accurate.

🧠 Stretching a wire: the n² rule

Stretch a wire to n times its length and its volume is unchanged, so the area becomes A/n. Then R = ρL/A becomes ρ(nL)/(A/n) = n²R — stretching to twice the length gives four times the resistance, and to three times gives nine. The same reasoning run backwards handles a wire drawn to half its length: R falls to a quarter. Candidates who change only L get 2R and 3R, and those are the options waiting for them.

Cells in Combination, and the Two Bridge Instruments

Cells in series: emfs add and internal resistances add, so I = nε/(R + nr) — useful when the external resistance is large.
Cells in parallel: the emf stays ε while the internal resistance falls to r/n, so I = ε/(R + r/n) — useful when the external resistance is small.
Meter bridge is a Wheatstone bridge with a uniform wire: at balance R/S = l/(100 − l) with l in centimetres from one end, and the unknown is read off the balance length.
Potentiometer compares two emfs as ε₁/ε₂ = l₁/l₂, and it is the accurate instrument precisely because it draws no current from the cell being measured at balance — which is why it measures emf where a voltmeter measures terminal voltage.

⚠️ Why the potentiometer beats the voltmeter

A voltmeter, however good, draws some current, so the cell it measures loses Ir internally and the reading is the terminal voltage rather than the emf. A potentiometer at balance draws zero current, so nothing is lost inside the cell and the reading is the emf itself. When a question asks why the potentiometer is preferred, the marked answer names the zero-current condition — not "because it is more sensitive" and not "because it has higher resistance", both of which are consequences at best.

Key takeaways

R = ρL/A, and ρ belongs to the material while R belongs to the piece. Stretching a wire to n times its length multiplies R by n², because the area falls by the same factor the length rises.
Series adds resistances with current common and gives more than the largest; parallel adds reciprocals with voltage common and gives less than the smallest. At constant V the smaller resistor dissipates more power, at constant I the larger.
With internal resistance, I = ε/(R + r) and terminal V = ε − Ir, so the terminal voltage equals the emf only on open circuit. Cells in series add emf and r; in parallel the emf stays and r falls to r/n.
A Wheatstone bridge is balanced when P/Q = R/S, and at balance the galvanometer’s own resistance stops mattering. A potentiometer measures emf rather than terminal voltage because at balance it draws no current.

Practice questions (8)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

A wire of resistance R is stretched uniformly to twice its original length. Its new resistance is:
1. 2R
2. 4R
3. R/2
4. R, since the material has not changed
Show answer
Answer: B — 4R
Stretching conserves volume, so doubling the length halves the cross-section, and R = ρL/A becomes ρ(2L)/(A/2) = 4R. Answering 2R changes the length and forgets the area, which is the commonest error in the topic and the reason it is offered first. R/2 changes the area in the wrong direction. And "R, since the material has not changed" confuses resistivity with resistance: ρ genuinely does not change, but R depends on the wire's dimensions as well, which is exactly the distinction the question is set on.
Three resistors of 6 Ω each are connected in parallel. The equivalent resistance is:
1. 18 Ω
2. 2 Ω
3. 3 Ω
4. 6 Ω
Show answer
Answer: B — 2 Ω
For n equal resistors in parallel the equivalent is R/n, so 6/3 = 2 Ω. The sanity check is that a parallel equivalent must be smaller than the smallest branch, which rules out 18 Ω (the series answer) and 6 Ω immediately. 3 Ω would be the answer for two such resistors rather than three, and it is the near-miss for a candidate who applies R/n with the wrong n. Reaching for the full reciprocal formula also works — 1/R = 1/6 + 1/6 + 1/6 = 1/2 — but the R/n shortcut is what the clock rewards.
A cell of emf 2 V and internal resistance 0.5 Ω is connected to a 1.5 Ω resistor. The terminal voltage is:
1. 2 V
2. 1.5 V
3. 1 V
4. 0.5 V
Show answer
Answer: B — 1.5 V
The current is I = ε/(R + r) = 2/(1.5 + 0.5) = 1 A, so the terminal voltage is V = ε − Ir = 2 − 1 × 0.5 = 1.5 V, which is also IR = 1 × 1.5. Answering 2 V reports the emf and ignores the internal drop, which is only correct on open circuit when no current flows. 1 V is the current in amperes read as a voltage, a units slip that the arithmetic invites here. And 0.5 V is the internal drop Ir itself, which is what has to be SUBTRACTED rather than reported.
Two bulbs rated 100 W and 60 W for the same mains voltage are connected in parallel to that supply. Which glows brighter, and why?
1. The 60 W bulb, because it has the higher resistance
2. The 100 W bulb, because at the same voltage P = V²/R and it has the lower resistance
3. Both equally, because they share the same voltage
4. The 60 W bulb, because it draws the larger current
Show answer
Answer: B — The 100 W bulb, because at the same voltage P = V²/R and it has the lower resistance
In parallel both bulbs get the full mains voltage, so P = V²/R applies with V fixed and the bulb of lower resistance dissipates more — the 100 W bulb, which is exactly what its rating promises at that voltage. The 60 W bulb does have the higher resistance, but at constant voltage that makes it dimmer, not brighter; the option states a true fact and draws the wrong conclusion from it. "Both equally" confuses equal voltage with equal power. And the 60 W bulb draws the smaller current, since I = V/R. The answer flips in SERIES, where the current is common and P = I²R makes the 60 W bulb brighter — that is the version worth remembering alongside this one.
In a balanced Wheatstone bridge with arms P, Q, R and S, the condition satisfied is:
1. P + Q = R + S
2. P/Q = R/S
3. PQ = RS
4. P/R = S/Q
Show answer
Answer: B — P/Q = R/S
Balance means no current through the galvanometer, which happens when the two arm ratios are equal: P/Q = R/S. Sums have no place in the condition, so the first option can be dismissed on form alone. PQ = RS is a product condition that does not follow from the ratio one — it is true only in the special case where the arms happen to be equal in pairs. P/R = S/Q rearranges the correct relation wrongly: cross-multiplying the true condition gives PS = QR, so the equivalent ratio form is P/R = Q/S rather than S/Q, and that swap is the trap here.
With rising temperature, the resistivity of a metallic conductor and of a semiconductor respectively:
1. Falls and falls
2. Rises and falls
3. Rises and rises
4. Falls and rises
Show answer
Answer: B — Rises and falls
In a metal the number of free electrons is essentially fixed, so heating only increases the collisions that impede them and the resistivity rises. In a semiconductor heating releases many more charge carriers across the gap, and that gain outweighs the extra scattering, so the resistivity falls — which is why a semiconductor has a negative temperature coefficient. "Falls and rises" is this answer exactly inverted and is the option to watch, since it is chosen by candidates who remember that the two behave oppositely but not which way round.
A potentiometer is preferred to a voltmeter for measuring the emf of a cell because:
1. It has a lower resistance than a voltmeter
2. At balance it draws no current from the cell, so no internal drop occurs
3. It can measure alternating as well as direct voltages
4. Its scale is longer, so the reading is easier to see
Show answer
Answer: B — At balance it draws no current from the cell, so no internal drop occurs
A voltmeter must draw some current to deflect, so the cell loses Ir internally and what is read is the terminal voltage, always a little less than the emf. At balance a potentiometer draws zero current, nothing is lost inside the cell, and the reading is the emf itself — a null method rather than a deflection method. Lower resistance is the reverse of what would help and is simply false here. The alternating-voltage claim is not a property of the basic potentiometer. And a longer scale improves precision of reading without touching the systematic error, which is the whole point being tested.
Why does a bulb light almost instantly when the switch is closed, even though the drift velocity of electrons is only about a millimetre per second?
1. Because the electrons in the filament travel at the speed of light
2. Because the electric field is set up along the whole circuit almost at the speed of light, and every electron starts drifting at once
3. Because the current in the filament is very large
4. Because the filament already stores electrons before the switch is closed
Show answer
Answer: B — Because the electric field is set up along the whole circuit almost at the speed of light, and every electron starts drifting at once
Closing the switch establishes the electric field throughout the conductor at nearly the speed of light, and the field is what sets every electron in the wire drifting simultaneously — including the ones already sitting in the filament. The light therefore appears at once even though any individual electron would take hours to travel from the switch to the bulb. Electrons moving at the speed of light contradicts the premise of the question and is impossible for a massive particle. A large current is neither necessary nor the mechanism. And the filament does contain free electrons already, which is true but explains nothing on its own: they only begin to drift when the field arrives, and the arrival of the field is the answer.