Exercises — Ohm's law — microscopic origin, resistivity

KYA chahiye: geometry se $R$ nikalna hai. YE formula kyun: $R=\rho L/A$ ek hi formula hai jo kisi material ke $\rho$ ko ek specific shape se jodata hai. $R=\rho \frac{L}{A}=\frac{(1.7\times 10^{-8})(3)}{2\times 10^{-6}}=\frac{5.1\times 10^{-8}}{2\times 10^{-6}}=0.0255\Omega .$ Answer: $R=2.55\times 10^{-2}\Omega$.

$\vec{J}=\sigma \vec{E}.$

• $\vec{J}$ = current density (A/m²), area ke per unit current.
• $\sigma$ = conductivity (S/m), $\sigma =n{e}^2\tau /m$.
• $\vec{E}$ = conductor ke andar electric field (V/m). Yeh law kehta hai ki response $\vec{J}$, cause $\vec{E}$ ke proportional hai, kyunki $\sigma$, $\vec{E}$ par depend nahi karta.

KYA: $R=\rho L/A$ ko $L$ ke liye solve karo. KYU: $R,\rho ,A$ teeno known hain. Pehle area convert karo: $A=0.4{\text{mm}}^2=0.4\times 10^{-6}{\text{m}}^2$. $L=\frac{RA}{\rho }=\frac{(22)(0.4\times 10^{-6})}{1.1\times 10^{-6}}=\frac{8.8\times 10^{-6}}{1.1\times 10^{-6}}=8\text{m}.$ Answer: $L=8\text{m}$.

KYA: $I=neA{v}_d$ ko $v_d$ ke liye solve karo. KYU: current wahi charge hai jo drifting carriers ek second mein sweep karte hain. $v_d=\frac{I}{neA}=\frac{5}{(1.8\times 10^{29})(1.6\times 10^{-19})(2\times 10^{-6})}.$ Denominator $=1.8\times 10^{29}\times 1.6\times 10^{-19}\times 2\times 10^{-6}=5.76\times 10^4$. $v_d=\frac{5}{5.76\times 10^4}\approx 8.68\times 10^{-5}\text{m/s}.$ Answer: $v_d\approx 8.7\times 10^{-5}\text{m/s}$ — hamesha ki tarah snail se bhi slow.

KYA: purane $R$ ke terms mein naya $R$ nikalo. KYU volume key constraint hai: stretching mein metal ki quantity conserve hoti hai, isliye $V=LA$ constant hai — agar $L$ double hoti hai, toh $A$ half honi chahiye. Maan lo purani wire: length $L$, area $A$, toh $R=\rho L/A$. Nayi wire: length $L^{\prime }=2L$. Volume constant $\Rightarrow L^{\prime }{A}^{\prime }=LA\Rightarrow A^{\prime }=\frac{LA}{2L}=\frac{A}{2}$. $R^{\prime }=\rho \frac{L^{\prime }}{A^{\prime }}=\rho \frac{2L}{A/2}=\rho \frac{4L}{A}=4R.$ Answer: resistance $4\times$ original ho jaata hai. Figure mein red nayi wire dekho: woh do guna lambi AUR aadhi moti hai, aur dono changes resistance badhate hain — isliye factor 4 hai, 2 nahi.

KYA: $v_d=I/(neA)$ use karke $v_d$ compare karo. KYU: same material matlab $n$ identical hai; $I$ bhi same hai — toh $v_d\propto 1/A$. Area $\propto$ diameter². Diameter double $\Rightarrow$ area $\times 4$. $\frac{v_{d,2}}{v_{d,1}}=\frac{A_1}{A_2}=\frac{1}{4}.$ Answer: Rod 2 ke electrons ek-chauthai speed se drift karte hain. Moti wire = zyada parallel lanes, toh same total current ke liye har electron slower chalta hai.

KYA aur KYU: geometry se $R$ nikalo, phir Joule heating formula mein daalo. (a) $R=\rho \frac{L}{A}=\frac{(1.1\times 10^{-6})(5)}{0.5\times 10^{-6}}=\frac{5.5\times 10^{-6}}{0.5\times 10^{-6}}=11\Omega .$ (b) $P=I^{2}R=(3{)}^2(11)=9\times 11=99\text{W}.$ Answers: $R=11\Omega$, $P=99\text{W}$.

KYA: $\rho =m/(n{e}^2\tau )$ ko invert karo. KYU: $\tau$ ke alawa sab known hai, aur $\tau$ woh microscopic quantity hai jo measurable $\rho$ ke andar chhupa hai. $\tau =\frac{m}{n{e}^2\rho }=\frac{9.1\times 10^{-31}}{(6.0\times 10^{28})(1.6\times 10^{-19}{)}^2(2.4\times 10^{-8})}.$ Denominator: $(1.6\times 10^{-19}{)}^2=2.56\times 10^{-38}$; times $6.0\times 10^{28}=1.536\times 10^{-9}$; times $2.4\times 10^{-8}=3.686\times 10^{-17}$. $\tau =\frac{9.1\times 10^{-31}}{3.686\times 10^{-17}}\approx 2.47\times 10^{-14}\text{s}.$ Answer: $\tau \approx 2.5\times 10^{-14}\text{s}$ (~25 femtoseconds), parent note ki copper value se consistent hai.

Part A — total resistance. Series mein resistances add hote hain kyunki same current dono se guzarta hai aur voltages stack hote hain. $R_1=\frac{(1.7\times 10^{-8})(2)}{1\times 10^{-6}}=3.4\times 10^{-2}\Omega ,R_2=\frac{(2.8\times 10^{-8})(3)}{2\times 10^{-6}}=4.2\times 10^{-2}\Omega .$ $R_{\text{tot}}=R_1+R_2=3.4\times 10^{-2}+4.2\times 10^{-2}=7.6\times 10^{-2}\Omega .$ Part B — field ratio. Field current se $J=\sigma E=E/\rho$ ke zariye juda hai, toh $E=\rho J=\rho I/A$. Kyunki $I$ shared hai: $\frac{E_1}{E_2}=\frac{{\rho }_1/A_1}{{\rho }_2/A_2}=\frac{{\rho }_1{A}_2}{{\rho }_2{A}_1}=\frac{(1.7\times 10^{-8})(2\times 10^{-6})}{(2.8\times 10^{-8})(1\times 10^{-6})}=\frac{3.4\times 10^{-14}}{2.8\times 10^{-14}}\approx \mathrm{1.214.}$ Answers: $R_{\text{tot}}\approx 7.6\times 10^{-2}\Omega$; $E_1/E_2\approx 1.21$. Dhyan do ki field ek series conductor ke along uniform nahi hoti — woh junction par jump karti hai (figure mein red boundary) kyunki wahan $\rho$ aur $A$ change hote hain, chahe $I$ na ho.

KYA: linear temperature law mein plug in karo. KYU yahan linear hai: is range mein $\tau$ roughly linearly $T$ ke saath girta hai, aur $\rho \propto 1/\tau$. $\Delta T=2020-20=2000\text{K}.$ $R=10[1+(4.5\times 10^{-3})(2000)]=10[1+9]=10\times 10=100\Omega .$ Answer: $R=100\Omega$ — das guna badha. Microscopic reason: garam lattice ions badi amplitude se vibrate karte hain, toh ek electron zyada baar collide karta hai → relaxation time $\tau$ girta hai → $\rho =m/(n{e}^2\tau )$ badhta hai → $R=\rho L/A$ badhta hai. Isliye hi ek lamp ka cold resistance chhota hota hai (switch-on par bada inrush current) lekin uska hot running resistance bada hota hai.

1.1 ::: $R=0.0255\Omega$ 2.1 ::: $L=8\text{m}$ 2.2 ::: $v_d\approx 8.7\times 10^{-5}\text{m/s}$ 3.1 ::: $R^{\prime }=4R$ 3.2 ::: $v_{d,2}/v_{d,1}=1/4$ 4.1 ::: $R=11\Omega$, $P=99\text{W}$ 4.2 ::: $\tau \approx 2.5\times 10^{-14}\text{s}$ 5.1 ::: $R_{\text{tot}}\approx 0.076\Omega$, $E_1/E_2\approx 1.21$ 5.2 ::: $R=100\Omega$