Class 12 Physics · Chapter 3
Short answer:
Is chapter me 19 exercise-style numericals cover kiye gaye hain — drift velocity, Ohm's law, resistivity/temperature-coefficient, series-parallel resistor combinations, EMF aur internal resistance, cells ki series/parallel combination, Kirchhoff's current aur voltage law (multi-loop circuit solving), aur Wheatstone bridge ka balance condition. Har numerical me given → formula → substitution → answer poore working steps ke saath diya gaya hai, taaki exam me full marks mile. Note: current 2026-27 rationalised syllabus me meter bridge aur potentiometer poori tarah drop ho chuke hain — is chapter me unka koi mention nahi hai.
Current Electricity chapter electron flow ko macroscopic quantities — current, resistance, EMF — se jodta hai, aur phir Kirchhoff's laws ke through complex circuits solve karna sikhata hai. Microscopic level pe free electrons random thermal motion me hote hain, lekin jab electric field lagta hai to un par ek chhota si average drift velocity superimpose ho jaati hai — yehi drift velocity current banati hai. Is chapter ka dusra bada hissa hai circuit analysis: series-parallel resistor combinations se shuru karke, EMF aur internal resistance wale real cells tak, aur end me Kirchhoff's junction rule aur loop rule jo kisi bhi complex multi-loop circuit ko systematically solve karne deते hain. Yeh chapter numerical-heavy hai — formula yaad karna kaafi nahi, circuit diagram padhna aur sahi sign convention lagana practice se aata hai.
Chapter 3 Summary — 5 Minute Revision
1. Electric current aur drift velocity
Conductor me free electrons random motion me hote hain (average velocity zero), lekin electric field E lagne par unpar ek chhota drift velocity vd superimpose hota hai, jiski direction field ke opposite hoti hai (electron negative hai).
- Current: I = nAvdq, jahan n = free electron density, A = cross-section area, q = electron charge
- Drift velocity chhoti hoti hai (~10-4 m/s order), lekin current turant establish ho jaata hai kyunki electric field wire me light-speed ke qareeb travel karta hai
- Conventional current ki direction electron drift ke opposite hoti hai
2. Ohm's law, resistivity aur conductivity
- Ohm's law: V = IR (constant temperature pe, ohmic conductors ke liye)
- Resistance: R = ρl/A — length ke proportional, area ke inversely proportional
- Resistivity ρ material property hai (geometry independent), conductivity σ = 1/ρ
- Metals: ρ temperature ke saath badhta hai. Semiconductors: ρ temperature ke saath ghatta hai
3. Temperature dependence of resistance
R = R0(1 + αΔT), jahan α = temperature coefficient of resistance, ΔT = temperature change
4. Series aur parallel combination of resistors
- Series: same current har resistor se guzarta hai, voltage divide hoti hai. Rs = R1 + R2 + ...
- Parallel: same voltage har resistor pe, current divide hota hai. 1/Rp = 1/R1 + 1/R2 + ...
- Parallel resistance hamesha smallest individual resistance se bhi kam hoti hai
5. EMF, internal resistance aur cell combinations
- Real cell: EMF (E) + internal resistance (r). Terminal voltage discharging me V = E − Ir
- Cells in series: Eeq = E1 + E2 + ..., req = r1 + r2 + ...
- Identical cells parallel me: Eeq = E (same), req = r/n (n cells)
6. Kirchhoff's laws
| Law | Statement | Physics basis |
|---|---|---|
| Junction rule (KCL) | Kisi junction pe ΣIin = ΣIout | Charge conservation — steady state me charge accumulate nahi ho sakta |
| Loop rule (KVL) | Kisi bhi closed loop me ΣΔV = 0 | Energy conservation — ek complete loop me electric potential energy conserve hoti hai |
↔ Table ko side me swipe karein
Kirchhoff's current law charge conservation se aata hai — kisi bhi junction par jitna current andar aata hai utna hi bahar jaana chahiye, warna charge kahin accumulate ho jayega jo steady state me ho hi nahi sakta. Loop rule energy conservation se aata hai — ek closed loop me ghoomkar wapas usi point pe aane par net potential change zero hona chahiye.
7. Wheatstone bridge (qualitative)
Char resistances P, Q, R, S ek bridge network me arranged hoti hain. Balance condition pe galvanometer me koi deflection nahi hota — is state me: P/Q = R/S. Yeh unknown resistance measure karne ka ek simple qualitative method hai.
Poore Class 12 Physics ke handwritten colour notes
IITian & district toppers ke banaye short notes — revision-ready, diagram ke saath. Board se pehle poora syllabus 3 din me revise.
Exercise Questions — Solutions (Q1–Q19)
Q1. Ek copper wire ka cross-sectional area 1×10-6 m² hai aur usme 1.5 A current flow ho raha hai. Free electron density n = 8.5×1028 per m³ hai. Drift velocity nikaalo.
Given: I = 1.5 A, A = 1×10-6 m², n = 8.5×1028 m-3, q = 1.6×10-19 C
vd = I / (nAq)
vd = 1.5 / (8.5×1028 × 1×10-6 × 1.6×10-19)
vd = 1.5 / (1.36×104) = 1.1×10-4 m/s
Answer: vd ≈ 1.1×10-4 m/s — itni chhoti drift velocity hone ke bawajood current turant establish ho jaata hai kyunki electric field wire me nearly light-speed se travel karta hai.
Q2. Ek resistor ka resistance 5 Ω hai aur usme 10 V potential difference hai. Current nikaalo.
Given: R = 5 Ω, V = 10 V
I = V/R = 10/5 = 2 A
Answer: I = 2 A
Q3. Ek wire ki length 2 m, area 0.5×10-6 m² aur resistance 4 Ω hai. Uska resistivity nikaalo.
Given: R = 4 Ω, l = 2 m, A = 0.5×10-6 m²
ρ = RA/l = (4 × 0.5×10-6) / 2
ρ = 1×10-6 Ω·m
Answer: ρ = 1×10-6 Ω·m
Q4. Ek wire ka resistance 20°C pe 10 Ω hai. Temperature coefficient α = 0.004 per °C hai. 100°C pe resistance nikaalo.
Given: R0 = 10 Ω, α = 0.004/°C, ΔT = 100 − 20 = 80°C
R = R0(1 + αΔT)
R = 10(1 + 0.004 × 80) = 10(1 + 0.32)
R = 10 × 1.32 = 13.2 Ω
Answer: R = 13.2 Ω
Q5. Teen resistors 2 Ω, 3 Ω, 5 Ω ek 20 V battery ke saath series me connect kiye gaye hain. Current aur har resistor ke across voltage nikaalo.
Given: R1=2Ω, R2=3Ω, R3=5Ω, V=20V, series combination
Rs = R1+R2+R3 = 2+3+5 = 10 Ω
I = V/Rs = 20/10 = 2 A
V1 = IR1 = 2×2 = 4 V, V2 = IR2 = 2×3 = 6 V, V3 = IR3 = 2×5 = 10 V
Answer: I = 2 A; V1=4V, V2=6V, V3=10V (check: 4+6+10=20V ✓)
Q6. Do resistors 6 Ω aur 3 Ω parallel me connected hain. Equivalent resistance nikaalo.
Given: R1=6Ω, R2=3Ω, parallel combination
1/Rp = 1/R1 + 1/R2 = 1/6 + 1/3 = 1/6 + 2/6 = 3/6
Rp = 2 Ω
Answer: Rp = 2 Ω — note ki yeh dono individual resistances se kam hai, jaisa parallel combination me hamesha hota hai.
Q7. Ek network me R1=4 Ω series me hai (R2=6 Ω parallel R3=3 Ω) ke saath. Total equivalent resistance nikaalo.
Given: R1=4Ω (series), R2=6Ω aur R3=3Ω (parallel to each other)
R23: 1/R23 = 1/6 + 1/3 = 1/6+2/6 = 3/6 ⇒ R23 = 2 Ω
Rtotal = R1 + R23 = 4 + 2 = 6 Ω
Answer: Rtotal = 6 Ω
Q8. Ek battery ki EMF 12 V hai aur internal resistance 1 Ω. Isse ek 5 Ω ka external resistor connect kiya gaya hai. Current aur terminal voltage nikaalo.
Given: E = 12 V, r = 1 Ω, R = 5 Ω
I = E/(R+r) = 12/(5+1) = 12/6 = 2 A
V = E − Ir = 12 − (2)(1) = 10 V
Answer: I = 2 A, terminal voltage V = 10 V (check: V = IR = 2×5 = 10V ✓)
Q9. Ek cell ko pehle 5 Ω resistor se connect karne par 0.5 A current milta hai, aur 11 Ω resistor se connect karne par 0.25 A current milta hai. Cell ki EMF aur internal resistance nikaalo.
Given: I1=0.5A with R1=5Ω; I2=0.25A with R2=11Ω
E = I1(R1+r) = I2(R2+r)
0.5(5+r) = 0.25(11+r)
2.5 + 0.5r = 2.75 + 0.25r ⇒ 0.25r = 0.25 ⇒ r = 1 Ω
E = 0.5(5+1) = 0.5 × 6 = 3 V
Answer: E = 3 V, r = 1 Ω (check: 0.25(11+1) = 0.25×12 = 3V ✓)
Q10. Teen identical cells, har ek ki EMF 1.5 V aur internal resistance 0.5 Ω, series me connect karke 4 Ω ke external resistor se joda gaya hai. Current nikaalo.
Given: 3 cells series, E each = 1.5V, r each = 0.5Ω, R = 4Ω
Etotal = 3 × 1.5 = 4.5 V
rtotal = 3 × 0.5 = 1.5 Ω
I = Etotal/(R + rtotal) = 4.5/(4+1.5) = 4.5/5.5
I ≈ 0.82 A
Answer: I ≈ 0.82 A
Q11. Do identical cells, har ek ki EMF 2 V aur internal resistance 1 Ω, parallel me connect karke 3 Ω ke external resistor se joda gaya hai. Current nikaalo.
Given: 2 identical cells parallel, E each = 2V, r each = 1Ω, R = 3Ω
Identical cells parallel me: Eeq = E = 2 V
req = r/n = 1/2 = 0.5 Ω
I = Eeq/(R + req) = 2/(3+0.5) = 2/3.5
I ≈ 0.57 A
Answer: I ≈ 0.57 A
Q12. Do cells — E1=4V (series me R1=2Ω ke saath) aur E2=2V (series me R2=2Ω ke saath) — ek common resistor R3=1Ω ke across parallel me connect kiye gaye hain. Kirchhoff's laws use karke I1, I2 aur I3 (R3 se guzarne wala current) nikaalo.
Given: E1=4V, R1=2Ω; E2=2V, R2=2Ω; common branch R3=1Ω. Maan lo I1 aur I2 apni branch se junction A ki taraf aate hain, aur I3 = I1+I2 R3 se guzarta hai (junction rule).
Junction rule: I3 = I1 + I2
Loop 1 (E1, R1, R3): E1 − I1R1 − I3R3 = 0 ⇒ 4 − 2I1 − I3 = 0
Loop 2 (E2, R2, R3): E2 − I2R2 − I3R3 = 0 ⇒ 2 − 2I2 − I3 = 0
Loop 1 se: I1 = (4 − I3)/2 = 2 − I3/2
Loop 2 se: I2 = (2 − I3)/2 = 1 − I3/2
I3 = I1+I2 = (2 − I3/2) + (1 − I3/2) = 3 − I3
2I3 = 3 ⇒ I3 = 1.5 A
I1 = 2 − 1.5/2 = 1.25 A, I2 = 1 − 1.5/2 = 0.25 A
Answer: I1 = 1.25 A, I2 = 0.25 A, I3 = 1.5 A
Verify: Loop 1: 4 − 2(1.25) − 1(1.5) = 4 − 2.5 − 1.5 = 0 ✓. Loop 2: 2 − 2(0.25) − 1(1.5) = 2 − 0.5 − 1.5 = 0 ✓. Junction: 1.25+0.25 = 1.5 ✓
Q13. Ek Wheatstone bridge me arms P=10 Ω, Q=15 Ω, R=20 Ω hain. Balance ke liye S ki value nikaalo.
Given: P=10Ω, Q=15Ω, R=20Ω, bridge balanced (no deflection in galvanometer)
Balance condition: P/Q = R/S
S = RQ/P = (20 × 15)/10 = 300/10
S = 30 Ω
Answer: S = 30 Ω
Q14. Ek balanced Wheatstone bridge me P=4 Ω, Q=6 Ω, R=8 Ω hain. S nikaalo.
Given: P=4Ω, Q=6Ω, R=8Ω, bridge balanced
P/Q = R/S ⇒ S = RQ/P
S = (8 × 6)/4 = 48/4
S = 12 Ω
Answer: S = 12 Ω
Q15. Ek battery (EMF=10V, internal resistance r=1 Ω) ek network se connect hai: R1=2 Ω series me hai (R2=6 Ω parallel R3=3 Ω) ke saath. Battery se total current, terminal voltage, aur R2 se guzarne wala current nikaalo.
Given: E=10V, r=1Ω, R1=2Ω series with (R2=6Ω || R3=3Ω)
R23: 1/R23 = 1/6+1/3 = 1/6+2/6 = 3/6 ⇒ R23 = 2 Ω
External R = R1+R23 = 2+2 = 4 Ω
Total circuit R = 4 + r = 4+1 = 5 Ω
I = E/Rtotal = 10/5 = 2 A
Terminal voltage V = E − Ir = 10 − 2(1) = 8 V
Voltage across parallel combo = V − IR1 = 8 − 2(2) = 4 V
I2 (through R2) = 4/6 = 0.67 A
Answer: I = 2 A, terminal voltage = 8 V, I2 ≈ 0.67 A (check: I3=4/3=1.33A, aur I2+I3=0.67+1.33=2A ✓)
Q16. Copper ki resistivity 1.7×10-8 Ω·m hai. Uski conductivity nikaalo.
Given: ρ = 1.7×10-8 Ω·m
σ = 1/ρ = 1/(1.7×10-8)
σ ≈ 5.88×107 S/m
Answer: σ ≈ 5.88×107 S/m
Q17. Ek copper wire ka resistance 0°C pe 4 Ω hai. Kis temperature pe uska resistance 4.8 Ω ho jayega, agar α = 0.004 per °C hai?
Given: R0=4Ω, R=4.8Ω, α=0.004/°C
R = R0(1+αT)
4.8 = 4(1 + 0.004T)
1.2 = 1 + 0.004T ⇒ 0.2 = 0.004T
T = 50°C
Answer: T = 50°C
Q18. Ek junction pe 4 wires milte hain. Do currents I1=3A aur I2=2A junction ke andar aa rahe hain, aur ek current I3=4A junction se bahar ja raha hai. Chautha current I4 kitna hoga (Kirchhoff's junction rule use karke)?
Given: I1=3A (in), I2=2A (in), I3=4A (out), I4 = ? (assume out)
ΣIin = ΣIout
3 + 2 = 4 + I4
I4 = 5 − 4 = 1 A
Answer: I4 = 1 A (junction se bahar) — yeh charge conservation ka direct consequence hai.
Q19. Ek resistor ka resistance 20 Ω hai aur usme 2 A current flow ho raha hai. Voltage aur power dissipated nikaalo.
Given: R = 20 Ω, I = 2 A
V = IR = 2 × 20 = 40 V
P = I²R = (2)² × 20 = 4 × 20 = 80 W
Answer: V = 40 V, P = 80 W
Important Equations — Ek Nazar Me
| Quantity | Formula |
|---|---|
| Current (drift velocity) | I = nAvdq |
| Drift velocity | vd = I/(nAq) |
| Ohm's law | V = IR |
| Resistance from resistivity | R = ρl/A |
| Conductivity | σ = 1/ρ |
| Temperature dependence of resistance | R = R0(1 + αΔT) |
| Resistors in series | Rs = R1 + R2 + ... |
| Resistors in parallel | 1/Rp = 1/R1 + 1/R2 + ... |
| Terminal voltage (discharging) | V = E − Ir |
| Terminal voltage (charging) | V = E + Ir |
| Cells in series | Eeq = E1+E2+..., req = r1+r2+... |
| Identical cells in parallel (n cells) | Eeq = E, req = r/n |
| Kirchhoff's junction rule (KCL) | ΣIin = ΣIout |
| Kirchhoff's loop rule (KVL) | ΣΔV = 0 (closed loop) |
| Wheatstone bridge balance | P/Q = R/S |
↔ Table ko side me swipe karein
Common Mistakes — Yahan Marks Kat te Hain
- Parallel resistances ko directly add kar dena. Rp = R1+R2 likhna galat hai — parallel me hamesha 1/Rp = 1/R1+1/R2 use karo, phir reciprocal lo.
- Kirchhoff loop equation me sign convention galat lagana. Current ki assumed direction ke saath move karte waqt resistor ka IR drop negative lena chahiye, EMF source ke + se − terminal ki taraf move karte waqt EMF negative lena chahiye — inconsistent sign lagana pura equation galat kar deta hai.
- EMF ko hi terminal voltage samajh lena. Jab current flow ho raha ho, terminal voltage V = E − Ir hota hai, E nahi — internal resistance ka Ir drop bhool jaana ek common galti hai.
- Negative current value aane par confuse ho jaana. Kirchhoff's equations solve karte waqt agar current negative aaye, iska matlab sirf itna hai ki actual current assumed direction ke opposite flow kar raha hai — magnitude sahi hoti hai.
- Drift velocity ki direction ko conventional current ki direction bata dena. Electrons negative charge carry karte hain, isliye unki drift velocity electric field ke opposite direction me hoti hai, jabki conventional current field ki direction me hi maana jaata hai — ye dono opposite hain.
- Wheatstone bridge me P, Q, R, S arms confuse kar dena. Balance condition P/Q = R/S me P-Q ek pair adjacent arms hain aur R-S dusra pair — bridge diagram me arms ko galat match karne se galat unknown resistance nikal aata hai.
Board-Style Important Questions
- 1 mark: Drift velocity ki definition dijiye aur bataiye iski direction conventional current se kaise related hai.
- 2 marks: Do resistors R1 aur R2 ko series aur phir parallel me connect karne par equivalent resistance ke formulas derive kijiye.
- 2 marks: Ek cell ki EMF aur terminal voltage me kya farq hai? Internal resistance ka role samjhaiye.
- 3 marks: Kirchhoff's junction rule aur loop rule ko state kijiye aur bataiye ye kaunse conservation laws se derive hote hain.
- 3 marks: Ek diye gaye circuit diagram me do EMF sources aur teen resistors hain — Kirchhoff's laws use karke sabhi branch currents nikalne ka procedure samjhaiye.
- 5 marks: Wheatstone bridge ka balance condition derive kijiye aur bataiye ise unknown resistance measure karne ke liye kaise use kiya jaata hai.
Aksar Poochhe Jaane Wale Sawaal
Drift velocity itni chhoti hoti hai to bulb turant kaise jal jaata hai?
Bulb turant isliye jalta hai kyunki electric field wire me nearly light-speed se travel karta hai, aur wire ke har electron par turant force lagti hai — sirf ek single electron ko wire ke ek sire se dusre sire tak pahunchne me time lagega, lekin sabhi electrons ek saath drift karna shuru kar dete hain, isliye current turant establish ho jaata hai.
Parallel combination me resistance kam kyun ho jaati hai?
Parallel me current ke paas multiple paths ho jaate hain, jo current flow karne ke liye effectively cross-section area badha deta hai — isliye total resistance har individual resistance se kam ho jaati hai.
Meter bridge aur potentiometer is chapter me kyun nahi hain?
Current 2026-27 rationalised NCERT syllabus me meter bridge aur potentiometer poori tarah drop kar diye gaye hain. Is chapter me sirf electric current, drift velocity, Ohm's law, resistivity, resistor combinations, EMF/internal resistance, cell combinations, Kirchhoff's laws aur Wheatstone bridge (qualitative) cover kiya gaya hai.
Kirchhoff's laws Ohm's law se better kyun hai complex circuits ke liye?
Ohm's law sirf simple series-parallel circuits ke liye kaam karta hai. Jab circuit me multiple EMF sources aur loops hon jo simple series-parallel me reduce nahi ho sakte, tab Kirchhoff's junction aur loop rules use karke systematically equations banakar sabhi unknown currents solve kiye ja sakte hain.
Cells ko series me lagana better hai ya parallel me?
Yeh depend karta hai requirement pe — series combination zyada EMF deti hai (voltage boost ke liye useful), jabki parallel combination internal resistance kam kar deti hai aur zyada current de sakti hai (jab external resistance internal resistance ke comparable ho, tab useful).
Wheatstone bridge balance hone ka matlab kya hai?
Balance ka matlab hai galvanometer me zero current flow ho raha hai — iska matlab bridge ke dono diagonal points same potential par hain. Is condition me P/Q = R/S hota hai, jo unknown resistance nikalne ke liye use hota hai.
Class 12 Physics — Saare Chapters
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