Question

5. Material with uniform resistivity is formed into a wedge as shown in Figure 27.19. Show that there sistance between face A and face of this wedges given by 6. In the circuit of Fig. 32-30, 1200 V, C-6,50 F, R.-R.-R.-7.30 x 10" n. With C completely unchanged the switch is suddenly closed t o la Determine the currents through each resistor for t=0 and 1 Draw qualitatively a graph of the potential drop V, across R. from O to tad What are the numerical values of V, att and told Give the physical meaning of and state a rough, but significant, bumerical lower bound, in seconds, for in this case. Fig. 31.30 L w 7. A circuit containing five resistors connected to a 12-V battery is shown in Fig. 28. What is the voltage drop across the 5.0-12 resistor? 6.00 120 500 8.(a) Find the equivalent capacitance between points a and b for the group of capacitors connected as shown in Figure 25.31 if C-5NF. C - 10 pF, and G- 2 F. NN w Fig. 28 8.(b) A copper wire of cross-sectional area 3 x 10-m ar- ries a current of 10 A. Find the drift velocity of the elec trons in this wire. The density of copper is 8.95 g/cm Figure 26.31 Two capacitors and C (where C > C) are charged to the same potential difference V. but with opposite po- larity. The charged capacitors are removed from the bat- tery, and their plates are connected as shown in Figure 26.10a. The switches S, and S, are then closed as in Figure 26.10b. (a) Find the final potential difference between a and b after the switches are closed. S 152 10. (a) Figure 26.10

Answer 1

Answer - 7(two images)

or 42 32 12 volt R, R, R. .. Resistance in series Resistance in parallel R = R1 + R₂ + R... RI so from the diagram circuit we can see that 3.0 & 54 Resistance in series sool 64 Parallel to each other so circuit can be change- un and is are in pagen - R 1+1 (Parallel) R Red +1 R-612 R 6 T R = 120 Ry - 10% = 40 32 su or Ry = R + R (series) Now circit can be drown. run u lserica MW 8e Parallel 4- = Int & un fr 42 I = V/R = I= 3 amp. 12% aap

by the diagram we can see that - I = It I = 3 amp Voltage in parallel combination is remain some so Voltage across 32 452 resistance is lovolt 7 8.2 I m I = 12% amp or 12 Volt/ohm 81 lav & Voltage across su reststance = Current through the resistance x tel Resi = I x 54 = 1 1985 = 7.5 Volt. 1 16 Voltage across 54 Resistance is 7.5 Volt

Answer - 8(b) (two images)

o drift velocity = Vd I= Current maa na no of electrons Per cubic meter A = Cross-section area q = charge of electrons " we have to find in first- " moles = m mass of copper - moler mass of Copper moss of capper - density & volume Volume = 1 m - 8.989 como mo or m= 8.95×10² kg/m3 m3 = 8,95x10 kg. & moler masss of copper = m = or m= 63.546 g/mol 0.0635 kg/mol moles = m. 8, 95% 16 kg - 140944.88 mo 0.0635 kg/mol i n = no. of e- in per meter cube of upper = moles y avogadro number = 140944.88 x 6.022 x 123 = 8.487 X 10 8 electronn / m3

-19 . I = 10 Amp. A = 3x10 m3, q = 1.6x10c 10 m/sec - Vol = 28 x 2 x 10 x 1.6*103 8,487810 0 x 3 x 15 x 16 x 101 - 6 Vd = 2.454 x 10 ml see

Answer- 8(a) (three images) I couldn't read the value of capacitor in figure- 26.31 . picture of diagram is not clear. so i have sovled the ideal problem which will help you to solve the exact problem.

@ @ In figure 26.31 The capacitor which is shown is not clearly visionable, but I have solve this by fig. It may help you in solve the exact problem. If you want to exact answer then e and the picture Jear, CS4 , H HH ③ Parallel combination - C = G+G+--- des combination - Serie in diagram ② & so b b c & C are ceg = + Ź connected in series, or Ceq - GQ G+

in diagram ① & ② ch are connected in series, G+ so cez = - + e or Ceq - Car so diagram can be replaced by new fig LILI GI J2) other Now Ceq , c & Cg in % are in parallel to each & c. 4 a are in 6 in in parallel to each other so for ② - I - Itt + 1 = 2 + 1

so again for 6 c'=CG + Ceq + G - 2 Ceq +G & for e c = c + Now diagram will be 9 Lc Now c'lch are in Capacitance will be series so equavalent Toto c' then send the pichure clearly, If you want more answer I will solve them.