Physics Class 12 Part I…
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Resistance Networks

🎓 Class 12 Physics CBSE Theory Ch 3 – Current Electricity ⏱ ~14 min
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આ MCQ મોડ્યુલ આના પર આધારિત છે: Resistance Networks

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Resistance Networks

3.7 Resistivity of Various Materials

Materials are classified into three broad categories based on resistivity (which spans more than 22 orders of magnitude!): conductors (\(\rho \approx 10^{-8}\) Ω·m), semiconductors (\(\rho \approx 10^{-5}\) to \(10^{6}\) Ω·m), and insulators (\(\rho \approx 10^{8}\) to \(10^{16}\) Ω·m).

Materialρ at 0 °C (Ω·m)α (K⁻¹)
Silver1.6 × 10⁻⁸+0.0041
Copper1.7 × 10⁻⁸+0.0068
Aluminium2.7 × 10⁻⁸+0.0043
Tungsten5.6 × 10⁻⁸+0.0045
Iron10 × 10⁻⁸+0.0065
Nichrome (alloy)~100 × 10⁻⁸+0.00017
Manganin (alloy)~48 × 10⁻⁸+0.000002
Carbon (graphite)3.5 × 10⁻⁵−0.0005
Germanium0.46−0.05
Silicon2300−0.07

3.8 Combinations of Resistors

In any circuit, resistors are combined in two basic ways: series (end-to-end) and parallel (across the same two nodes). Most networks reduce to these two with repeated application.

3.8.1 Resistors in Series

When several resistors are joined end-to-end so that the same current flows through each, they are said to be in series. The total potential drop is the sum:

\[ V = V_1 + V_2 + V_3 = I R_1 + I R_2 + I R_3 \quad\Rightarrow\quad R_\text{eq} = R_1 + R_2 + R_3 \]
V R₁ R₂ R₃ I
Fig. 3.4: Three resistors in series. Same current I flows through each; voltages add. R_eq = R₁ + R₂ + R₃.

3.8.2 Resistors in Parallel

When several resistors are joined between the same two nodes so that they share the same potential difference, they are said to be in parallel. The total current is the sum of the branch currents:

\[ I = I_1+I_2+I_3 = \dfrac{V}{R_1}+\dfrac{V}{R_2}+\dfrac{V}{R_3} \quad\Rightarrow\quad \dfrac{1}{R_\text{eq}} = \dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}\]
V R₁ R₂ R₃ I₁ I₂ I₃ I →
Fig. 3.5: Three resistors in parallel. Same V across each; currents add. 1/R_eq = 1/R₁ + 1/R₂ + 1/R₃.
Quick check: Two equal resistors R in series give 2R; in parallel give R/2. So three equal R's: series → 3R; parallel → R/3.

Interactive Simulation: Series & Parallel Calculator L3 Apply

Choose three resistor values and the connection type. The equivalent resistance and total current update live.

Equivalent resistance R_eq = 60.00 Ω
Total current I = V/R_eq = 0.20 A
Total power P = VI = 2.40 W

3.9 Color Code of Carbon Resistors

Commercial carbon resistors are too small to write the value on. Instead, the value is encoded in four colored bands painted around the body. Bands 1–2 give the first two significant digits, band 3 is the multiplier (power of 10), and band 4 is the tolerance.

ColorDigit (Band 1, 2)Multiplier (Band 3)Tolerance (Band 4)
Black010⁰ = 1—
Brown110¹—
Red210²—
Orange310³—
Yellow410⁴—
Green510⁵—
Blue610⁶—
Violet710⁷—
Grey810⁸—
White910⁹—
Gold—10⁻¹±5%
Silver—10⁻²±10%
No band——±20%
Yellow Violet Red Gold 4 7 ×10² ±5% Value: 47 × 10² Ω = 4.7 kΩ ± 5%
Fig. 3.6: Decoding a 4-band carbon resistor: yellow-violet-red-gold = 4700 Ω ± 5%.

Interactive Simulation: Resistor Color Code Decoder L3 Apply

Pick the four bands of a resistor and read the value instantly.

Resistance = 4700 Ω = 4.7 kΩ ± 5%

3.10 Temperature Dependence of Resistivity

Resistivity of metals increases with temperature; for semiconductors it decreases. Over a limited range,

\[ \rho_T = \rho_0 \big[1 + \alpha(T - T_0)\big] \]

where \(\alpha\) is the temperature coefficient of resistivity.

  • Metals (α > 0): n is roughly constant; rising T increases lattice vibrations, decreases τ, so ρ rises.
  • Semiconductors (α < 0): rising T frees more carriers (n increases sharply), overcoming the τ decrease — ρ falls.
  • Alloys (manganin, constantan, nichrome): very small α — used as standard resistors and heating coils.
(a) Metal — ρ rises with T T ρ (b) Semiconductor — ρ falls with T T ρ (c) Nichrome — ρ ≈ constant T ρ
Fig. 3.7: ρ–T behaviour for (a) metal, (b) semiconductor, (c) alloy with small α.
Superconductivity: For some metals (e.g. mercury) ρ drops abruptly to zero below a critical temperature T_c (~4 K for Hg). Such materials carry currents indefinitely without resistance — discovered by Kamerlingh Onnes in 1911.

Worked Example 1: Series-Parallel Network

In the network shown: R₁ = 4 Ω, R₂ = 6 Ω in parallel, then in series with R₃ = 3 Ω. Battery V = 9 V. Find (a) R_eq, (b) total current, (c) current through R₂.

(a) R₁∥R₂ = (4×6)/(4+6) = 24/10 = 2.4 Ω. R_eq = 2.4 + 3 = 5.4 Ω.
(b) I = V/R_eq = 9/5.4 ≈ 1.67 A.
(c) Voltage across parallel pair = I × 2.4 = 4.0 V. So I₂ = 4.0/6 ≈ 0.67 A.

Worked Example 2: Color Code

A resistor has color bands: Brown – Black – Orange – Silver. Find its value and tolerance.

Brown=1, Black=0, Orange=×10³, Silver=±10%.
Value = 10 × 10³ Ω = 10 kΩ ± 10%.
Range: 9 kΩ to 11 kΩ.

Worked Example 3: Temperature Effect

A platinum resistor reads 5.00 Ω at 27 °C and 5.795 Ω at the steady operating temperature of a furnace. Take α = 3.92 × 10⁻³ K⁻¹. Estimate the temperature of the furnace.

\[ R_T = R_0[1+\alpha(T-T_0)] \] \[ \dfrac{R_T}{R_0}-1 = \alpha(T-T_0) \] \[ \dfrac{5.795}{5.00}-1 = 0.159 = 3.92\times10^{-3}(T-27) \] \[ T-27 = 40.6 \;\Rightarrow\; T \approx \boxed{67.6\ °C} \] (For an actual furnace one would expect a much larger ΔR; this is a low-temperature illustration.)
Activity 3.2 — Series vs Parallel BulbsL4 Analyse

Materials: two identical 6 V/0.5 A bulbs, a 6 V battery, switch, wires.

Procedure: First connect the two bulbs in series across the battery and observe their brightness. Then connect them in parallel and compare.

Predict: Which arrangement will make each bulb glow more brightly? Why?

Observation: In series, each bulb gets only 3 V, so each is dim. In parallel, each bulb gets the full 6 V and glows at full brightness.

Conclusion: Domestic appliances are connected in parallel so each receives the full mains voltage (220 V) and so that switching one off does not interrupt the others.

Competency-Based Questions

A laboratory resistor box has three resistors of 4 Ω, 6 Ω and 12 Ω. A student combines them in different ways and measures the equivalent resistance with a multimeter. She also examines color-coded carbon resistors and notices how their values change when warmed.

Q1. The minimum equivalent resistance achievable with the three resistors is: L3 Apply

  • (a) 22 Ω (all in series)
  • (b) 4 Ω
  • (c) 2 Ω (all in parallel)
  • (d) 12 Ω
(c) 1/R = 1/4+1/6+1/12 = 3/12+2/12+1/12 = 6/12 = 1/2, so R_eq = 2 Ω. Parallel always gives the minimum.

Q2. Short Answer: Why does the resistance of a metal increase with temperature, but that of a semiconductor decrease? L4 Analyse

In a metal, n is essentially constant; rising T increases lattice vibrations, decreasing τ, so ρ ∝ 1/τ rises. In a semiconductor, the dominant effect is the thermal generation of charge carriers — n increases exponentially with T, swamping the small drop in τ, so ρ falls.

Q3. A carbon resistor has bands Green-Blue-Brown-Gold. Its value is: L3 Apply

  • (a) 56 Ω ± 5%
  • (b) 560 Ω ± 5%
  • (c) 5.6 kΩ ± 5%
  • (d) 5.6 Ω ± 10%
(b) Green=5, Blue=6, Brown=×10, Gold=±5% → 56 × 10 = 560 Ω ± 5%.

Q4. Fill in the blanks: Manganin and nichrome have very ____ temperature coefficients of resistivity, which makes them suitable for ____. L2 Understand

small (≈ zero); standard resistors and heating elements (since their resistance hardly changes with temperature).

Q5. HOT: Two wires of the same material and same length but cross-sectional areas in the ratio 1:2 are first joined in series and then in parallel. Find the ratio of their equivalent resistances. L6 Create

Let R₁ = R, then R₂ = R/2 (area is doubled).
Series: R + R/2 = 3R/2.
Parallel: (R · R/2)/(R + R/2) = (R²/2)/(3R/2) = R/3.
Ratio = (3R/2) : (R/3) = 9 : 2.

Assertion–Reason Questions

Choose: (A) Both A and R true; R explains A. (B) Both true; R does not explain A. (C) A true, R false. (D) A false, R true.

Assertion (A): When resistors are connected in parallel, the equivalent resistance is always less than the smallest individual resistance.

Reason (R): In parallel, the reciprocals of resistances add up.

(A) Both true and R correctly explains A. Adding 1/R terms makes 1/R_eq larger, hence R_eq smaller than each individual R.

Assertion (A): Heating elements (e.g. toaster, iron) are made of nichrome, not copper.

Reason (R): Nichrome has a high resistivity and a high melting point.

(A) Both true and R explains A. Copper would melt and offers too little resistance to dissipate heat.

Assertion (A): The temperature coefficient α of a semiconductor is negative.

Reason (R): The number of free charge carriers in a semiconductor decreases with increasing temperature.

(C) A is true, but R is FALSE — the number of free carriers in a semiconductor INCREASES with T, which is precisely why ρ decreases (α negative).

Frequently Asked Questions - Resistance Networks

What is the main concept covered in Resistance Networks?
In NCERT Class 12 Physics Chapter 3 (Current Electricity), "Resistance Networks" covers the core principles and equations students need for board exam success. The MyAiSchool lesson explains the topic with definitions, derivations, worked examples, and interactive simulations. Key formulas and dimensional analysis are included to build conceptual depth and problem-solving skills aligned with the CBSE 2025-26 syllabus.
How is Resistance Networks useful in real-life applications?
Real-life applications of "Resistance Networks" from NCERT Class 12 Physics Chapter 3 include electronics, communication systems, medical imaging, solar energy, semiconductor devices, and modern technology. The MyAiSchool lesson links every concept to a tangible example so students see physics as a problem-solving framework for the physical world, not as abstract formulas.
What are the key formulas in Resistance Networks?
Key formulas in "Resistance Networks" (NCERT Class 12 Physics Chapter 3 Current Electricity) are derived step-by-step in the MyAiSchool lesson. Students should memorize the final formula AND understand its derivation for full board marks. Each formula is listed with its dimensional formula, SI unit, applicability range, and common pitfalls. The Summary section at the end of each part includes a quick-reference formula card.
How does this part connect to other parts of Chapter 3?
NCERT Class 12 Physics Chapter 3 (Current Electricity) is structured so each part builds on the previous one. "Resistance Networks" connects directly to neighbouring parts via shared definitions, units, and methodology. The MyAiSchool lesson cross-references related concepts with internal links so students can navigate the whole chapter as one connected story rather than disconnected fragments.
What types of CBSE board questions come from Resistance Networks?
CBSE board questions from "Resistance Networks" typically include: (1) 1-mark MCQs on definitions and formulas, (2) 2-mark short-answer derivations or applications, (3) 3-mark numerical problems with units, (4) 5-mark long-answer derivations followed by application. The MyAiSchool lesson tags each Competency-Based Question (CBQ) with Bloom level (L1-L6) so students know how to study for each weight.
How can students use the interactive simulation effectively?
The interactive simulation in the "Resistance Networks" lesson allows students to adjust input parameters (sliders or selectors) and see physical quantities update in real time. To use it effectively: (1) try extreme values to understand limiting cases, (2) compare with the analytical formula, (3) check unit consistency, (4) test special configurations from worked examples. The simulation reinforces conceptual intuition that pure formula manipulation cannot.
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Physics Class 12 Part I – NCERT (2025-26)
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