V = integral^integral_i E. Ds i_e = n_eAv_d v_d = eT/m E I = dQ/dt J = I/A I = d
ID: 1451714 • Letter: V
Question
V = integral^integral_i E. Ds i_e = n_eAv_d v_d = eT/m E I = dQ/dt J = I/A I = delta V/R J = sigma E sigma = n_e e^2 r/m^e p = 1/sigma R = p L/A R_ser = R_1 + R_2 + R_N R_par = (1/R_1 + 1/R_2 + ... 1/R_N)^-1 P_bat = I epsilon P_R = I delta V_R Q = Q_0e^-t/T t = RC A 2.0-mm-diameter copper wire and a 1.0-mm-diameter aluminum wire are joined into a single wire as shown in Fig. (1). A current flows through the wire. The electric field in the copper section of the wire is E_Cu = 0.010 V/m. What is the electric field Em in the aluminum section of the wire? Data: sigma_Cu = 6.0 times 10^7 omega^-1 m^-1, sigma_Al = 3.5 times 10^7 omgea^-1 m^-1 In the circuit shown in Fig. (2), describe what happens to the brightness of bulbs A and B when the switch is closed. Explain. Find the current through and the potential across each resistor in the circuit shown in Fig. (3). Find the rate that energy is supplied by the battery. Consider the circuit in Fig. (4) and assume the filament bulb is Ohmic with R = 50 Omega. The capacitor is initially uncharged when the switch is closed. Make a sketch of the brightness of the bulb as a function of time. Clearly justify your graph including a discussion of Kirchoff's Voltage Law. A long time after the switch is closed, how much power is dissipated in the light bulb? After the switch has been closed for a long time it is reopened. Make a sketch of the brightness of the bulb as a function of time after the switch is opened. Clearly justify your graph. Reconsider parts (a) and (c). Which if either reaches it's final state faster and why?Explanation / Answer
you can see in the figure that both wire is in series combination.
Now remember that in series comb. current distribution remains same.
Current in copper wire = current in aluminum wire
J = i/A
J = sigma*E
from both equation
E = i/(A*sigma)
Now,
iCu = iAl
which gives us
EAl/Ecu = (sigmaCu*Acu)/(sigmaAl*Aal)
EAl = Ecu*sigmaCu*dCu^2/(sigmaAl*dAL^2)
EAl = 0.010*2^26*10^7/(3.5*10^7*1.0^2)
EAl = 0.0685 V/m
Let me know if you have any doubt.
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.