Worked examples — Ohm's law — microscopic origin, resistivity

1.8.16 · D3 · Physics › Electromagnetism › Ohm's law — microscopic origin, resistivity

Yeh page parent note pe build karta hai aur Electric current and current density, Drude model of conduction, Temperature dependence of resistance, Joule heating — power dissipation, aur Resistors in series and parallel se connect karta hai.

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Scenario matrix

Kuch bhi solve karne se pehle, chalte hain list karte hain is topic ke har alag tarah ke problem ko. Neeche har worked example ek cell fill karta hai.

Cell	Kya cheez isse alag banati hai	Covered by
A. Basic forward	Geometry + $\rho$ diya, $R$ nikalo	Ex 1
B. Basic inverse	$R$, geometry diya, $\rho$ nikalo (material ID)	Ex 2
C. Micro chain	$E$ (ya $V$) se → $v_d$ → $J$ → $I$	Ex 3
D. Scaling / ratios	"$L$ double karo", "radius halve karo" — numbers ki zaroorat nahi	Ex 4
E. Limiting / degenerate	$\tau \to 0$, $n\to 0$, $A\to 0$, insulator/superconductor	Ex 5
F. Temperature case	$\tau$ change hota hai $T$ ke saath; change ka sign	Ex 6
G. Real-world word problem	Ghar ki wire garam hona	Ex 7
H. Exam twist	Do wires, same material, series/parallel combo	Ex 8

Yahan signs kabhi negative nahi hote (kyunki $n,e^2,\tau ,m,L,A$ sab positive hain), isliye trig-style problems ke "quadrant" cases ki jagah yahan limiting cases hain: kya hota hai jab koi quantity $0$ ya $\infty$ ki taraf jaaye. Cell E inhe deliberately dhundta hai.

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Setup — jo constants baar baar aayenge

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Cell A — Basic forward

Step 1 — Radius ko metres mein convert karo aur area nikalo. $r=0.2\text{mm}=0.2\times 10^{-3}\text{m}=2\times 10^{-4}\text{m}$ $A=\pi r^2=\pi (2\times 10^{-4}{)}^2=\pi \times 4\times 10^{-8}\approx 1.257\times 10^{-7}{\text{m}}^2$ Yeh step kyun? $R=\rho L/A$ mein area chahiye, aur wire round hai, isliye $A=\pi r^2$. Millimetres ko metres banana zaroori hai warna powers of ten galat ho jayenge.

Step 2 — $R=\rho L/A$ mein daalo. $R=\frac{(5.6\times 10^{-8})(3)}{1.257\times 10^{-7}}\approx 1.34\Omega$ Yeh step kyun? Yeh parent note se master geometry formula hai.

Verify: Units: $\frac{\Omega \text{m}\cdot \text{m}}{{\text{m}}^2}=\Omega$. ✓ Order of magnitude: numerator $\sim 10^{-7}$, denominator $\sim 10^{-7}$, toh $R\sim 1\Omega$. Forecast "kuch ohms" se match karta hai. ✓

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Cell B — Basic inverse (material identify karo)

Step 1 — Square cross-section ka area. $A=(2\times 10^{-3})(2\times 10^{-3})=4\times 10^{-6}{\text{m}}^2$ Yeh step kyun? Cross-section square hai, isliye $A={\text{side}}^2$; pehle mm→m convert karo.

Step 2 — $R=\rho L/A$ ko invert karke $\rho$ nikalo. $\rho =\frac{RA}{L}=\frac{(2.125\times 10^{-3})(4\times 10^{-6})}{0.5}=1.7\times 10^{-8}\Omega \text{m}$ Yeh step kyun? $R$ diya hai aur $\rho$ chahiye, isliye rearrange karo — dono sides ko $A/L$ se multiply karo.

Verify: $1.7\times 10^{-8}$ bilkul copper ki resistivity hai → rod copper ki hai. Units: $\frac{\Omega \cdot {\text{m}}^2}{\text{m}}=\Omega \text{m}$. ✓

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Cell C — Poori microscopic chain

Step 1 — $v_d=eE\tau /m$ se drift speed nikalo. $v_d=\frac{(1.6\times 10^{-19})(0.1)(8\times 10^{-15})}{9.1\times 10^{-31}}\approx 1.41\times 10^{-4}\text{m/s}$ Yeh step kyun? Parent ka Step 3 result: field ek free-flight $\tau$ ke liye push karta hai usse pehle ki collision sab reset kar de. Figure dekho — ulajha hua path thermal wandering hai; woh tiny rightward slope $v_d$ hai.

Step 2 — Current density $J=ne{v}_d$. $J=(1.8\times 10^{29})(1.6\times 10^{-19})(1.406\times 10^{-4})\approx 4.05\times 10^6{\text{A/m}}^2$ Yeh step kyun? Count karo kitna charge per second per unit area sweep hota hai — parent ka Step 4.

Step 3 — Current $I=JA$. $I=(4.05\times 10^6)(10^{-6})\approx 4.05\text{A}$ Yeh step kyun? Current = current-density poore cross-section pe faili hui.

Verify: $\sigma$ se cross-check: $\sigma =n{e}^2\tau /m=(1.8\times 10^{29})(1.6\times 10^{-19}{)}^2(8\times 10^{-15})/(9.1\times 10^{-31})\approx 4.05\times 10^7\text{S/m}$, toh $J=\sigma E=4.05\times 10^6$. ✓ Do tarikhon se same answer. Drift speed $\sim 10^{-4}\text{m/s}$ — ek kachuwe ki chaal, jaisa forecast tha. ✓

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Cell D — Scaling aur ratios (numbers ki zaroorat nahi)

Step 1 — (a) Length double, area fixed. $R=\rho \frac{L}{A}\Rightarrow R\propto L\Rightarrow R_a=2R_0$ Yeh step kyun? $\rho$ aur $A$ unchanged hain, isliye $R$ seedha $L$ ke saath scale karta hai — series mein double collisions.

Step 2 — (b) Radius halve, length fixed. $A=\pi r^2\Rightarrow r\to r/2⟹A\to A/4⟹R_b=\frac{R_0}{1/4}=4R_0$ Yeh step kyun? $R\propto 1/A$ aur area $r^2$ ke saath jaata hai, isliye $r$ halve karne se $A$ quarter hoti hai aur $R$ quadruple ho jaata hai.

Step 3 — (c) Same volume mein double length pe redraw. Volume $V=AL$ fixed hai. Agar $L\to 2L$ toh $A\to A/2$. Toh $R_c=\rho \frac{2L}{A/2}=4\rho \frac{L}{A}=4R_0$ Yeh step kyun? Metal ka fixed lump stretch karne se woh dono lambi aur patli ho jaati hai. Dono effects $R$ ko upar push karte hain, giving a factor $2\times 2=4$.

Verify: Ranking: (a) $2\times$, (b) $4\times$, (c) $4\times$. Thinning dominate karta hai. Ek general redraw rule: fixed volume pe $R\propto L^2$ — double length → $4\times$. ✓

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Cell E — Limiting aur degenerate cases

Step 1 — (i) $\tau \to \infty$. $\rho =\frac{m}{n{e}^2\tau }\to \frac{m}{n{e}^2\cdot \infty }=0\Rightarrow R\to 0$ Yeh step kyun? Koi collision nahi matlab brake gayab — electrons freely drift karte hain, resistivity khatam. Yeh idealised superconductor limit hai.

Step 2 — (ii) $n\to 0$. $\rho =\frac{m}{n{e}^2\tau }\to \frac{m}{0}=\infty \Rightarrow R\to \infty$ Yeh step kyun? Koi free carriers nahi, koi current possible nahi — perfect insulator. Isliye $n$ (na ki $\tau$) metal aur insulator ke beech ka switch hai (dekho Semiconductors vs metals).

Step 3 — (iii) $A\to 0$. $R=\rho \frac{L}{A}\to \rho \frac{L}{0}=\infty$ Yeh step kyun? Saari current ek vanishing cross-section se squeeze karo → infinite resistance. Physically yahi woh jagah hai jahan ek patla filament overheat karke blow karta hai (link to Joule heating in Ex 7).

Step 4 — (iv) $L\to 0$. $R=\rho \frac{L}{A}\to \rho \frac{0}{A}=0$ Yeh step kyun? Travel karne ki koi length nahi → koi accumulated collisions nahi → zero resistance. Ek wide, flat slab almost ek short circuit hai.

Verify: $R=0$ tak pahunchne ke do tarike (perfect crystal, ya zero length), $R=\infty$ tak pahunchne ke do tarike (no carriers, ya zero area). Material limits ($\tau ,n$) $\rho$ mein rehte hain; shape limits ($A,L$) geometry factor mein rehte hain. ✓

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Cell F — Temperature case

Step 1 — $R$ ki $\tau$ pe dependence likho. $R=\rho \frac{L}{A}=\frac{m}{n{e}^2\tau }\cdot \frac{L}{A}\Rightarrow R\propto \frac{1}{\tau }$ Yeh step kyun? Geometry aur $n$ fixed hain, toh sirf $\tau$ move karta hai — aur woh denominator mein hai.

Step 2 — Ratio lo. $\frac{R_{\text{hot}}}{R_{\text{cold}}}=\frac{{\tau }_{\text{cold}}}{{\tau }_{\text{hot}}}=\frac{2.5\times 10^{-14}}{2.0\times 10^{-14}}=1.25$ Yeh step kyun? Ratios se saare shared constants ($m,n,e,L,A$) cancel ho jaate hain, sirf do $\tau$'s bachte hain.

Verify: $R$ 25% badhta hai. Sign check: chhota $\tau$ → bada $1/\tau$ → bada $R$. Metals garam hone pe zyada resistive hote hain, bilkul jaisa Temperature dependence of resistance mein hai. ✓

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Cell G — Real-world word problem

Step 1 — (a) Geometry se Resistance. $A=0.3{\text{mm}}^2=0.3\times 10^{-6}{\text{m}}^2$ $R=\rho \frac{L}{A}=\frac{(1.1\times 10^{-6})(1.5)}{0.3\times 10^{-6}}=5.5\Omega$ Yeh step kyun? Koi bhi current ya power se pehle $R$ chahiye — yeh object ki fixed property hai.

Step 2 — (b) Ohm's law se Current. $I=\frac{V}{R}=\frac{230}{5.5}\approx 41.8\text{A}$ Yeh step kyun? Macroscopic $V=IR$, rearranged. (Ek real toaster longer/thinner coil use karta hai kam current kheenchne ke liye; yeh simplified wala deliberately zyada kheenchta hai.)

Step 3 — (c) Joule heating se Power. $P=VI=(230)(41.8)\approx 9.6\times 10^3\text{W}$ Yeh step kyun? Power = voltage × current — electrical energy ka har joule element mein heat ban jaata hai (dekho Joule heating — power dissipation).

Verify: Cross-check $P=V^2/R=230^2/5.5=52900/5.5\approx 9.6\times 10^3\text{W}$. ✓ Dono tarikhon se same. Units: $\text{V}\cdot \text{A}=\text{W}$. ✓

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Cell H — Exam twist (combination)

Step 1 — Radius change se $R_2$ nikalo. Twice the radius → $A_2=\pi (2r{)}^2=4\pi r^2=4A_1$. Kyunki $R\propto 1/A$ (same $\rho$, $L$): $R_2=\frac{R_1}{4}=\frac{3}{4}=0.75\Omega$ Yeh step kyun? Sirf area differ karta hai; area radius squared ke saath scale karta hai, isliye $R$ 4 se ghatta hai.

Step 2 — (a) Series. $R_{\text{series}}=R_1+R_2=3+0.75=3.75\Omega$ Yeh step kyun? Series mein current dono se baari baari guzarti hai, isliye resistances add hote hain (parent note: "collisions in series").

Step 3 — (b) Parallel. $\frac{1}{R_{\text{par}}}=\frac{1}{3}+\frac{1}{0.75}=0.3333+1.3333=1.6667{\Omega }^{-1}$ $R_{\text{par}}=\frac{1}{1.6667}=0.6\Omega$ Yeh step kyun? Parallel paths conductances add karte hain; zyada parallel routes → kam total resistance (dekho Resistors in series and parallel).

Verify: Sanity rule: parallel total ($0.6\Omega$) sabse chhoti branch ($0.75\Omega$) se kam hai. ✓ Series total ($3.75\Omega$) sabse badi branch ($3\Omega$) se zyada hai. ✓ Dono consistency checks pass; aur $0.6<1$, forecast confirm. ✓

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Active Recall

Recall Kaun se cell mein

$A=\pi r^2$ chahiye aur kaun se mein $A={\text{side}}^2$? Round wires mein $A=\pi r^2$ (Ex 1, 3, 8); square/rectangular cross-sections mein side×side (Ex 2).

Recall Fixed volume mein wire ko double length pe redraw karna

$R$ ko kitne se multiply karta hai… $4$ — kyunki constant volume pe $R\propto L^2$ (Ex 4c).

Recall Kaun si quantity metal↔insulator flip karti hai,

$\tau$ ya $n$? $n$: $n\to 0$ se $\rho \to \infty$ (insulator); $\tau$ sirf tune karta hai ki ek metal kitna achha conductor hai (Ex 5).

Recall Garam metal:

$\tau$ $R$ ko kya karta hai? $\tau$ girta hai, aur kyunki $R\propto 1/\tau$, $R$ badhta hai (Ex 6).

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Connections

• Ohm's law — microscopic origin, resistivity (index 1.8.16) (parent)
• Electric current and current density
• Drude model of conduction
• Resistors in series and parallel
• Temperature dependence of resistance
• Joule heating — power dissipation
• Semiconductors vs metals