Current Electricity

Current ElectricityNEET Physics · Class 12 · NCERT Chapter 3

Overview Notes NEET Questions Interactive Learning

7 interactive concept widgets for Current Electricity. Drag any slider, change any number, and watch the formula and the answer update live. Built so you understand how each NEET problem actually works, not just the final number.

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Ohm's law solver

Resistance from resistivity

Electrical power and energy

Series and parallel resistors

EMF and internal resistance

Kirchhoff loop solver

Wheatstone bridge balance

Ohm's law and resistivity

V = I R for any unknown, plus how resistance depends on shape and material.

Ohm's law

Ohm's law solver (V = I R)

Pick the unknown, enter the other two, get the answer.

Ohm's law: V = I R. Pick the unknown and enter the other two.

V (V)

I (A)

R (Ω)

Voltage V (V)

12.000

V = I R

Try this

Doubling V at fixed R doubles I. Doubling R at fixed V halves I.
Power: P = VI = I²R = V²/R. Use whichever the problem hands you.
Ohm's law works for ohmic materials (most metals). Diodes and transistors are non-ohmic.

Resistivity

Resistance from resistivity

R = rho L over A. Try different materials and dimensions.

Resistance is bigger if the wire is longer or thinner. Resistivity rho is the material's contribution.

Selected: Copper, ρ = 1.68e-8 Ω·m

Length L: 1.00 m

Cross-section A: 1.00 mm²

Resistance

16.800 mΩ

R = \frac{ρ L}{A}

Try this

Copper has very low resistivity, which is why most wiring uses it.
Silicon is a semiconductor: many orders of magnitude more resistive than metals.
Doubling length doubles R. Doubling area halves R.
Resistance is a property of the wire; resistivity is a property of the material.

Power

Electrical power and energy

P = V I = I² R = V² / R. Three forms of the same equation.

Power dissipated by a resistor. NEET often gives bulb wattage and asks for resistance, or gives R and V and asks for P.

Voltage V: 220 V (Indian mains: 220 V)

Resistance R: 484 Ω

Time on: 5 hours

Power

100.0 W

Current I = 0.455 A

Energy used

0.500 kWh

1 kWh costs about ₹6 to ₹8 in India

P = V I = I^{2} R = \frac{V ^{2}}{R}

Try this

A 100 W bulb on 220 V mains has R = V²/P = 484 Ω.
Two 100 W bulbs in series: each gets less voltage, so each emits less power. Total combined < 100 W.
Two 100 W bulbs in parallel: each at full voltage, each at 100 W. Total = 200 W.
1 unit of electricity = 1 kWh. A 1 kW heater run for 1 hour = 1 unit.

Combinations and EMF

Resistors in series and parallel, plus the role of internal resistance in real cells.

Combinations

Resistors in series and parallel

The opposite of capacitor combinations. Series adds R; parallel adds reciprocals.

Three resistors. Series: total adds up. Parallel: reciprocals add.

R₁: 2 Ω

R₂: 3 Ω

R₃: 6 Ω

Effective resistance

11.000 Ω

R = R_{1} + R_{2} + R_{3}

Try this

Series: same current through all. Voltages add. Total R is bigger than any single one.
Parallel: same voltage across all. Currents add. Total R is smaller than the smallest.
Two equal resistors R in parallel: equivalent is R/2.

EMF and r

EMF, internal resistance and terminal voltage

A real cell has internal resistance. The terminal voltage drops below the EMF as current flows.

Real cell: EMF (epsilon) plus internal resistance (r). When current flows, terminal voltage V drops below the EMF.

EMF ε: 12.0 V

Internal resistance r: 0.50 Ω

External R: 5.0 Ω

Terminal voltage

10.91 V

V = ϵ - I r

2.182 A

P_ext

23.80 W

P_lost (in r)

2.38 W

Try this

When R_ext is much greater than r, terminal voltage is close to ε (cell is barely loaded).
Short-circuit (R_ext → 0): I = ε/r, terminal V → 0, all power dissipates inside.
Maximum power transfer to load: when R_ext = r. Then half the power is lost inside.
A new cell has r ≈ 0.1 Ω; an old cell can have r > 1 Ω, which is why dim torches mean drained batteries.

Kirchhoff and Wheatstone

Two cells sharing a load, plus the bridge that NEET asks about every year.

Kirchhoff

Kirchhoff loop solver (two cells, one load)

Two cells in parallel sharing a load. Computed via Kirchhoff's laws and equivalent-cell shortcut.

Two cells (ε₁, r₁) and (ε₂, r₂) in parallel driving a common load R. Solved using Kirchhoff's laws.

ε₁: 12 V, r₁: 1.00 Ω

ε₂: 6 V, r₂: 2.00 Ω

Load R: 5.00 Ω

Equivalent cell

ε_eq = 10.000 V, r_eq = 0.667 Ω

I_load

1.765 A

I₁

3.176 A

I₂

-1.412 A

ϵ_{eq} = \frac{ϵ _{1} r _{2} + ϵ _{2} r _{1}}{r _{1} + r _{2}}

Try this

If ε₁ ≠ ε₂, current can flow from one cell to the other even with no external load (a "wasted" current).
Equal cells in parallel: ε_eq = ε, r_eq = r/2. Current capacity doubles.
I₁ negative means cell 1 is being charged by cell 2, so it acts like a load.

Wheatstone bridge

Wheatstone bridge balance

The classic NEET circuit. Four resistors in a diamond, plus a galvanometer. Balance when P/Q = R/S.

Wheatstone bridge with four resistors. The bridge is balanced (no current through the galvanometer) when P/Q = R/S.

P: 10 Ω

Q: 20 Ω

R: 30 Ω

S: 60 Ω

Battery V: 12 V

Bridge state

BALANCED

P/Q = 0.500, R/S = 0.500

V_B − V_D = 0.000 V

Balanced when \frac{P}{Q} = \frac{R}{S}

Try this

Set P = R and Q = S. The bridge balances. (Or any pair where P/Q = R/S.)
Balance condition does not depend on the battery voltage or galvanometer resistance.
Used to find an unknown resistance: balance and read off the unknown from S = QR/P.

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