Current Electricity | Class 12 CBSE Physics | Worked Numericals
The three formulas, and which one to reach for
Electrical power dissipated in a resistor can be written three equivalent ways:
P = VI P = I²R P = V²/R
These aren’t three different facts — they’re the same fact (P=VI combined with Ohm’s law, V=IR) rearranged in three ways. The skill this note builds isn’t memorizing all three — it’s picking the one that avoids an unnecessary extra calculation step, given what a problem actually hands you.
Quick rule: if you’re given I and R, use P=I²R directly. If you’re given V and R, use P=V²/R directly. Only reach for P=VI when you’re given (or can immediately read off) BOTH V and I — using it when you only have I and R (or only V and R) forces you to compute the missing third quantity first, which is extra work and an extra place to make an arithmetic slip.
Worked Example 1 — direct substitution, no shortcuts needed
A 100 Ω resistor carries a current of 0.5 A. Find the power dissipated.
Solution: Given I and R directly → use P = I²R:
P = (0.5)² × 100 = 0.25 × 100 = 25 W
Verification using a different formula: First find V = IR = 0.5×100 = 50V, then P = V²/R = 50²/100 = 2500/100 = 25 W. ✓ Matches, but notice this took an extra step — exactly the inefficiency the quick rule above is warning against.
Worked Example 2 — the “wrong formula, extra step, and a place to slip” trap
A student is given I = 2A and R = 15Ω, and asked to find power. They choose to use P = VI, so they first calculate V = IR = 30V, then P = VI = 2×30 = 60W. Is the final answer correct, and what’s the issue with the method?
Answer: The final numerical answer (60 W) is correct — but the method invites unnecessary risk. Since I and R were given directly, P = I²R = 4×15 = 60W gets the same answer in one step, with one less opportunity for an arithmetic error (in this case computing V) to creep in. This is a “process” issue, not a “wrong answer” issue — but on a timed exam, or on a problem with more components, an extra unnecessary step compounds risk. Choosing the right form of the power formula, given what’s directly provided, is itself a tested skill — not just a matter of taste.
Worked Example 3 — series circuit: same current, different power
A 12V battery drives two resistors, 4Ω and 8Ω, in series. Find the power dissipated in each resistor.
Solution: In series, the SAME current flows through both resistors. First find the total current:
R_total = 4+8 = 12Ω
I = V/R_total = 12/12 = 1A
Now use P = I²R for each resistor individually (same I, different R):
P₄ = I²×4 = 1²×4 = 4W
P₈ = I²×8 = 1²×8 = 8W
Notice: the larger resistor (8Ω) dissipates MORE power here, not less — because in a series circuit, current is fixed and P = I²R means larger R directly gives larger P. This is the opposite of what many students expect if they’ve absorbed “more resistance = less power” as a blanket rule without noticing it only holds for constant voltage, not constant current.
Worked Example 4 — parallel circuit: same voltage, the pattern flips
The same two resistors, 4Ω and 8Ω, are instead connected in parallel across the same 12V battery. Find the power dissipated in each.
Solution: In parallel, the SAME voltage (12V) appears across both resistors. Use P = V²/R for each:
P₄ = 12²/4 = 144/4 = 36W
P₈ = 12²/8 = 144/8 = 18W
Notice: now the SMALLER resistor (4Ω) dissipates MORE power — the exact opposite ranking from the series case, using the identical pair of resistors and the identical battery. This directly confirms Worked Example 3’s warning: whether “more resistance → more power” or “more resistance → less power” depends entirely on whether current or voltage is being held fixed by the circuit configuration, not on some universal rule about resistance and power.
Worked Example 5 — energy, not just power
A 1000 W (1 kW) heater is run for 3 hours. Find the electrical energy consumed, in both joules and kilowatt-hours.
Solution: Energy = Power × time.
In joules (SI, use time in seconds):
E = 1000 W × (3×3600 s) = 1000×10800 = 1.08×10⁷ J
In kWh (the unit your electricity bill actually uses):
E = 1 kW × 3 h = 3 kWh
Both describe the same physical energy — kWh is simply a more convenient unit for household-scale consumption than joules, since 1 kWh = 3.6×10⁶ J is a much more “human-sized” number for a day’s appliance usage than the equivalent joule figure.
Common errors to watch for
| Error | What’s actually true |
|---|---|
| Using P=VI when only I and R (or only V and R) are given | Requires an unnecessary extra step to find the missing quantity first; use P=I²R or P=V²/R directly instead |
| Assuming “higher resistance always means lower power,” universally | Only true when voltage is held constant (parallel/independent branches). For constant current (series), higher resistance means HIGHER power |
| Confusing power (rate, in watts) with energy (total amount, in joules or kWh) | Energy = Power × time; a high-power device run briefly can consume less total energy than a low-power device run for a long time |
| Forgetting to convert time to seconds when computing energy in joules | E=Pt requires t in seconds if P is in watts and E is wanted in joules; using hours directly gives kWh instead — both are valid, but they must not be mixed |
Quick Practice
Q1. A resistor has 20V across it and draws 4A. Find the power dissipated (use the most direct formula — you’re given both V and I).
Answer: P = VI = 20×4 = 80W
Q2. Two identical resistors, R each, are first connected in series and then in parallel across the same battery of voltage V. In which configuration does the combination dissipate more total power?
Answer: Parallel. Series equivalent resistance is 2R, so total power = V²/(2R). Parallel equivalent resistance is R/2, so total power = V²/(R/2) = 2V²/R — four times larger than the series case. Lower total resistance (parallel) draws more total current from the same voltage, and dissipates more total power.
Q3. A 60W bulb and a 100W bulb (both rated for the same 220V mains supply) are compared. Which one has the higher resistance, and how do you know from the ratings alone?
Answer: The 60W bulb has HIGHER resistance. Using P=V²/R with the same V for both: R=V²/P, so R is inversely proportional to P at fixed voltage. The lower-power bulb (60W) has the higher resistance; the higher-power bulb (100W) has the lower resistance — this often surprises students who expect “more powerful” to mean “more resistance.”
Summary
| Given | Use directly | Avoid |
|---|---|---|
| V and I | P = VI | Computing R first, unless R is separately needed |
| I and R | P = I²R | P = VI (requires finding V first) |
| V and R | P = V²/R | P = VI (requires finding I first) |
| Circuit type | What’s fixed | Effect of larger R on that component’s power |
|---|---|---|
| Series | Current | Larger R → MORE power (P=I²R) |
| Parallel | Voltage | Larger R → LESS power (P=V²/R) |
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