In a SERIES circuit the current is the same in every resistor, so use P = I²R: the LARGER resistor dissipates MORE power. In a PARALLEL circuit the voltage is the same across every resistor, so use P = V²/R: the SMALLER resistor dissipates MORE power. Memory hook: "Series → I same → bigger R wins; Parallel → V same → smaller R wins."
Series (top): same current, so P = I²R makes the 200 ohm resistor dissipate twice the power of the 100 ohm. Parallel (bottom): same voltage, so P = V²/R makes the smaller 100 ohm resistor dissipate more and get hotter.
Your doubts, answered
In series, which resistor dissipates more power - large or small?
In series the current I is the SAME through both resistors. Use P = I²R. Since I is common, power is directly proportional to R. So the LARGER resistance dissipates MORE power. Example: 100 ohm and 200 ohm in series carry the same I, and P(200)/P(100) = 200/100 = 2, so the 200 ohm gives twice the heat.
In parallel, which resistor dissipates more power?
In parallel the voltage V is the SAME across both resistors. Use P = V²/R. Since V is common, power is inversely proportional to R. So the SMALLER resistance dissipates MORE power (it draws more current). This is the exact opposite of the series case, which is why students mix them up.
When do I use I²R and when do I use V²/R?
Pick the formula that keeps the SHARED quantity fixed. Series shares current, so P = I²R is cleanest (P proportional to R). Parallel shares voltage, so P = V²/R is cleanest (P inversely proportional to R). All three forms P = VI = I²R = V²/R are always true; you just choose the one that avoids extra algebra.
For the SAME resistors on the SAME supply, does series or parallel dissipate more total power?
Parallel dissipates MORE total power. Total power from a fixed source is P = V²/R_eq. Parallel gives a SMALLER equivalent resistance than series, so V²/R_eq is larger. That is why bulbs shine brighter in parallel (household wiring) and dimmer in series (old fairy lights).
Two identical bulbs: are they brighter in series or in parallel?
Brighter in parallel. Each bulb in parallel gets the full supply voltage V, so each dissipates V²/R. In series the voltage splits, so each bulb gets only V/2 and dissipates (V/2)²/R = V²/4R - one quarter of the parallel power per bulb. Full voltage across each bulb means maximum brightness.
⚠️ The NEET trap ✗ Using P = V²/R for a series circuit and concluding the small resistor gets hot, or using P = I²R for parallel and picking the large resistor. ✓ Match the formula to the shared quantity. Series → same I → P = I²R → LARGER R dissipates more. Parallel → same V → P = V²/R → SMALLER R dissipates more. 🧠 Same current, bigger R wins. Same voltage, smaller R wins. Write the shared quantity first, then choose the formula.
Real NEET questions
2024
Two heaters A and B rated 1 kW and 2 kW are connected first in series and then in parallel to a fixed source. The ratio of power outputs (series : parallel) is:
A · 2 : 9 ✓
B · 1 : 2
C · 2 : 3
D · 1 : 1
Solution: Step 1 - find the resistances. At the rated voltage V, P = V²/R, so R = V²/P. Thus R_A = V²/1 and R_B = V²/2 (in units where V² = 1, R_A = 1, R_B = 0.5). Resistance is inversely proportional to rated power. Step 2 - series total power. R_series = R_A + R_B = 1 + 0.5 = 1.5. P_series = V²/R_series = 1/1.5 = 2/3. Step 3 - parallel total power. In parallel the total power is the SUM of individual rated powers = 1 + 2 = 3 (each gets full V). So P_parallel = 3. Step 4 - ratio. P_series : P_parallel = (2/3) : 3 = 2 : 9. Answer: A.
2022
Two resistors 100 ohm and 200 ohm are connected in parallel. The ratio of thermal energy developed in 100 ohm to that in 200 ohm in a given time is:
A · 1 : 2
B · 2 : 1 ✓
C · 1 : 4
D · 4 : 1
Solution: Step 1 - identify the shared quantity. In PARALLEL the voltage V is the same across both resistors. Step 2 - choose the right formula. With V common, heat H = Pt = (V²/R)t, so H is inversely proportional to R. Step 3 - take the ratio. H(100)/H(200) = R(200)/R(100) = 200/100 = 2. So H(100) : H(200) = 2 : 1 - the SMALLER 100 ohm resistor develops MORE heat. Answer: B.
2019
Six similar bulbs are connected with a DC source of emf E and zero internal resistance in two sections A and B, each section having 3 bulbs in parallel and the two sections in series. The ratio of power consumed when (i) all six glow to (ii) when two bulbs of section A and one bulb of section B glow, is:
A · 4 : 9
B · 9 : 4 ✓
C · 1 : 2
D · 2 : 1
Solution: Let each bulb have resistance R. Case (i) all glow: section A = 3 in parallel = R/3, section B = R/3, in series gives R_eq = R/3 + R/3 = 2R/3. Power P_i = E²/R_eq = E²/(2R/3) = 3E²/(2R). Case (ii): section A has 2 in parallel = R/2, section B has 1 bulb = R, in series gives R_eq = R/2 + R = 3R/2. Power P_ii = E²/(3R/2) = 2E²/(3R). Ratio P_i : P_ii = (3/2) : (2/3) = 9 : 4. Answer: B. This uses P = E²/R_eq: smaller total resistance draws more power.
Solved Current Electricity NEET PYQs
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Is total power the same in series and parallel for the same resistors?
No. On the same fixed supply, parallel dissipates more total power because it has a smaller equivalent resistance and P = V²/R_eq. Series has a larger R_eq, so less total power.
Why does the smaller resistor heat up more in parallel?
In parallel every resistor gets the full supply voltage. With V fixed, P = V²/R, so a smaller R draws more current and dissipates more power - it gets hotter.
Why does the larger resistor heat up more in series?
In series the same current flows through all resistors. With I fixed, P = I²R, so a larger R drops more voltage and dissipates more power.
Which is the safest single formula to remember?
All three are equal: P = VI = I²R = V²/R. For series use I²R (I is shared); for parallel use V²/R (V is shared). Matching the formula to the shared quantity avoids mistakes.
Are home appliances wired in series or parallel?
Parallel. Each appliance gets the full 220 V and runs at its rated power, and one device switching off does not cut the others. Series would split the voltage and dim everything.
