Understanding Resistance, Drift Velocity, and Capacitor Charge

School

Worcester Polytechnic Institute**We aren't endorsed by this school

Course

PHYSICS 1120

Subject

Electrical Engineering

Date

Dec 11, 2024

Pages

Uploaded by MagistrateHerring4704

25.34.Identify andSet Up:The resistance is the same in both cases, and ? = 𝑉2/?.Execute:(a)Solving ? = 𝑉2/?for R, gives ? = 𝑉2/?.Since the resistance is the same in both cases, we have 𝑉12𝑃1=𝑉22𝑃2.Solving for P2givesP2= P1(V2/V1)2= (0.0625 W)[(12.5 V)/(1.50 V)]2= 4.41 W.(b)Solving for V2gives 𝑉2= 𝑉1√?2?1= (1.50 V)√5.00 W0.0625 W= 13.4 V.Evaluate:These calculations are correct assuming that the resistor obeys Ohm’s law throughout the range of currents involved.25.2.IDENTIFY:𝐼 = ?/?.Use 𝐼 = ?|𝑞|𝑣𝑑𝐴to calculate the drift velocity d.vSET UP:? = 5.8 × 1028m−3.|𝑞| = 1.60 × 10−19C.EXECUTE:(a)𝐼 =??=420 𝐶80(60 ?)= 8.75 × 10−2A.(b)𝐼 = ?|𝑞|𝑣𝑑𝐴.This gives 𝑣𝑑=𝐼?|𝑞|𝐴=8.75 × 10−2𝐴(5.8 × 1028)(1.60 × 10−19𝐶)(𝜋(1.3 × 10−3?)2)= 1.78 × 10−6m/s.EVALUATE:𝑣𝑑is smaller than in Example 25.1, because Iis smaller in this problem.

24.30.IDENTIFY:This problem involves dielectrics and capacitors in parallel.SET UP:First sketch the circuit as in Fig. 24.30. Ceq= C1+ C2, C = KC0, and Q = CV. We want the charge in both cases.Figure 24.30 EXECUTE:(a)Q = CeqV = (C1+ C2)V= (10.0 µF)(36.0 V) = 360 µC.(b)C1is now KC0= (5.00)(4.00 µF) = 20.0 µF. Q = CeqV= (26.0 µF)(36.0 V) = 936 µC.EVALUATE:The total charge increases due to the insertion of the dielectric. The dielectric increases the equivalent capacitance which increases the stored charge.