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Q.An aluminium wire of diameter 0.24 cm is connected in series with a copper wire of diameter 0.16 cm. If a 10 ampere current is flowing through it, find — (2+2=4)

(i) Current density in the aluminium wire.
(ii) Drift velocity of electrons in the copper wire.
(Given: In copper, number of electrons per cubic metre volume =8.4×1028=8.4\times10^{28})
(OR)
(i) A 900 pF capacitor is charged by a 100 V battery. How much electrostatic energy is stored by the capacitor? [1½]
[FIGURE: A 900 pF capacitor C connected across a 100 V battery, with its plates carrying charge +Q+Q and −Q-Q.]
(ii) The capacitor is disconnected from the battery and connected to another 900 pF capacitor. What is the electrostatic energy stored by the system now? [1½]
[FIGURE: Two 900 pF capacitors C connected together (in parallel), each shown with charged plates.]
(iii) Is the energy the same in both cases? If not, why? [1]
Uttarakhand UbseUttarakhand Board Intermediate (Class 12) 2024Subjective· 4mImportance★★★★★
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Current density J=I/AJ=I/A; drift velocity vd=I/(nAe)v_d=I/(nAe) — both need the actual cross-sectional area of the respective wire.

  1. Current density in the aluminium wire. Diameter =0.24 cm⇒=0.24\ \text{cm}\Rightarrow radius r=0.12 cm=1.2×10−3 mr=0.12\ \text{cm}=1.2\times10^{-3}\ \text{m}. AAl=πr2=π(1.2×10−3)2=4.52×10−6 m2A_{Al}=\pi r^2=\pi(1.2\times10^{-3})^2=4.52\times10^{-6}\ \text{m}^2 JAl=IAAl=104.52×10−6≈2.21×106 A/m2J_{Al}=\frac{I}{A_{Al}}=\frac{10}{4.52\times10^{-6}}\approx2.21\times10^{6}\ \text{A/m}^2
  2. Drift velocity of electrons in the copper wire. Since the wires are in series, the same current I=10 AI=10\ \text{A} flows through the copper wire. Diameter =0.16 cm⇒=0.16\ \text{cm}\Rightarrow radius r=0.08 cm=8×10−4 mr=0.08\ \text{cm}=8\times10^{-4}\ \text{m}. ACu=πr2=π(8×10−4)2≈2.01×10−6 m2A_{Cu}=\pi r^2=\pi(8\times10^{-4})^2\approx2.01\times10^{-6}\ \text{m}^2 Using I=nAevdI=nAev_d, with n=8.4×1028 m−3n=8.4\times10^{28}\ \text{m}^{-3}, e=1.6×10−19 Ce=1.6\times10^{-19}\ \text{C}: vd=InACue=10(8.4×1028)(2.01×10−6)(1.6×10−19)≈3.70×10−4 m/sv_d=\frac{I}{nA_{Cu}e}=\frac{10}{(8.4\times10^{28})(2.01\times10^{-6})(1.6\times10^{-19})}\approx3.70\times10^{-4}\ \text{m/s}

OR — Capacitor energy.

(i) Energy stored =12CV2=12(900×10−12)(100)2=12(900×10−12)(104)=4.5×10−6 J=4.5 μJ=\dfrac{1}{2}CV^2=\dfrac{1}{2}(900\times10^{-12})(100)^2=\dfrac{1}{2}(900\times10^{-12})(10^4)=4.5\times10^{-6}\ \text{J}=4.5\ \mu\text{J}.

(ii) Initial charge on the first capacitor: Q=CV=(900×10−12)(100)=9×10−8 CQ=CV=(900\times10^{-12})(100)=9\times10^{-8}\ \text{C}. When disconnected from the battery and connected to an identical uncharged 900 pF900\ \text{pF} capacitor (in parallel, so both have equal capacitance), the total charge QQ redistributes equally between the two identical capacitors, and the common potential becomes:

V′=QC1+C2=9×10−81800×10−12=50 VV'=\frac{Q}{C_1+C_2}=\frac{9\times10^{-8}}{1800\times10^{-12}}=50\ \text{V}

Energy now stored in the system:

U2=12(C1+C2)V′2=12(1800×10−12)(50)2=12(1800×10−12)(2500)=2.25×10−6 J=2.25 μJU_2=\frac{1}{2}(C_1+C_2)V'^2=\frac{1}{2}(1800\times10^{-12})(50)^2=\frac{1}{2}(1800\times10^{-12})(2500)=2.25\times10^{-6}\ \text{J}=2.25\ \mu\text{J}

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