In which one among the following situations, the bulb ⊗ would glow the most ? (Consider all batteries are the same)
- (a)One cell driving two bulbs connected in series
- (b)Two cells in series driving two bulbs connected in series
- (c)One cell driving a single bulb
- (d)Two cells in series driving two bulbs connected in parallel
Answer
Why
Correct — D. Brightness is the power dissipated in a bulb, so compare P = V²/R for one bulb in each printed circuit. Write each cell as EMF E and each bulb as resistance R.
(a) one cell, two bulbs in series: the two bulbs share E, so each gets E/2 and P = E²/4R.
(b) two cells, two bulbs in series: total 2E shared between two bulbs, so each gets E and P = E²/R.
(c) one cell, one bulb: the bulb gets the full E, P = E²/R.
(d) two cells, two bulbs in parallel: each bulb sits directly across the full 2E, so P = 4E²/R.
Parallel is the key. Bulbs in parallel do not divide the voltage between them — each one receives the whole supply. Combine that with the doubled EMF and (d) delivers four times the power of (b) or (c), and sixteen times that of (a).
Why the others are wrong
- (a)One cell driving two bulbs connected in series — The dimmest of the four. A single cell's EMF is split between two bulbs in series, so each receives only E/2 and dissipates a quarter of the power a single bulb on one cell would.
- (b)Two cells in series driving two bulbs connected in series — Doubling the cells and then splitting the voltage between two series bulbs returns each bulb to E — the same as a single bulb on a single cell, and four times less than the parallel arrangement.
- (c)One cell driving a single bulb — The bulb gets the full E, which beats both series arrangements but is still only a quarter of what each bulb receives in (d), where two cells drive bulbs in parallel.
Concept
Cells in series add their EMFs. Bulbs in series share the supply voltage between them, so each gets a fraction; bulbs in parallel each sit across the full supply voltage. Since power in a fixed resistance goes as the square of the voltage across it, doubling a bulb's voltage quadruples its brightness.
Two independent choices are being tested at once — how many cells, and how the bulbs are wired — and the wiring matters more because of the square. Going from one cell to two doubles the voltage and so quadruples power; going from series bulbs to parallel bulbs doubles it again for each bulb. The option that wins is the one that gets both right, which is why (d) is far ahead rather than marginally so.
Key facts
- Cells in series: EMFs add.
- Bulbs in series share the supply voltage; bulbs in parallel each get all of it.
- Power in a bulb P = V^2/R, so voltage counts twice over.
- (a) E^2/4R, (b) E^2/R, (c) E^2/R, (d) 4E^2/R.
- Adding bulbs in parallel does not dim the others, unlike adding them in series.
Most EMF and no voltage-sharing — option (d).
Study next
Common traps
- Assuming more bulbs always means dimmer, regardless of wiring.
- Forgetting that power depends on the SQUARE of the voltage.
- Treating parallel bulbs as sharing the supply voltage.
NDA prints several battery-and-bulb circuits and asks which glows most — count the cells for the supply voltage, then check whether the bulbs share it or each receive it in full.
Related PYQs
Which of the following arrangement of resistors offers minimum effective resistance between points X and Y ?
- (a) [circuit diagram: three 3Ω resistors connected in parallel between X and Y]
- (b) [circuit diagram: two 3Ω resistors connected in parallel between X and Y]
- (c) [circuit diagram: two 1Ω resistors connected in series between X and Y]
- (d) [circuit diagram: three 1Ω resistors connected in series between X and Y]
Answer(a) [circuit diagram: three 3Ω resistors connected in parallel between X and Y]
The same series-versus-parallel contrast on the resistance side — there parallel branches drive the total resistance down, here they drive each bulb's voltage, and so its brightness, up.
Practice
- practice — not a real PYQ
Two identical bulbs are connected in parallel across a battery. If one bulb is removed, the other bulb
- (a)glows brighter
- (b)glows dimmer
- (c)glows with the same brightness
- (d)goes out
Answer(c) glows with the same brightness — in parallel each bulb keeps the full supply voltage. - practice — not a real PYQ
If the voltage across a bulb is doubled, the power it dissipates becomes
- (a)half
- (b)double
- (c)four times
- (d)unchanged
Answer(c) four times — P = V²/R, so power goes as the square of the voltage.