Question

I4=？

The emfs in the figure below are 8 - 5.00 V and 8 - 24.0 V. The resistances are Ri = 17.00, R2 - 32.00, R3 = 48.00, and R4 = 58.00. Find the magnitude of the current in each resistor when the switch is in the following states. R2 w RA R w R (a) open 110247 12 - 0 247 130247 14 = 0 ✓ A (b) closed 11 - 0 25534 12 - 0245 ✓ A 13 - 0245 А 00138 x Your rescose differs from the correct answer by more than 109 Double check your calculations. A

Answer 1

@ When the switch is open: Ra Loop D wwwwww Coop Rua Ri &=5v E = 24 V R,= 172 R₂ = 322 R3 = 48 R4 = 582 0 + S & S R3 When the switch is open current in loop 1 must be zero ie i=o. Otherwise Kirchoff's cunnent law will be violated. Applying KVL in loop a we get Raig + R, (litig) + Raiz - En (Ryt Rgt Rg) iz & (: 1,=0) => (17+32+48) ig = 24 24/97 0.24742 -- ñ 0.247

are So current through the resistons R., Rg, Rg and R4 respectively 1, = 1,t ig=/0+ 0.247) A = 0.247 A 12 = {q = 0.247 in = 0.247 13 - 14 ا 0 A

(bWhen the switch is closed :- Ra WWWWWW Loop 1) (Loop 2 i Ri is En + + www R3 Applying & Let the current in loop. I and loop 2 are resp respectively į and is (as shown in Big bique) KVL in loop s, we get Rs (i tig) + Ry = - 17 i+1)+5% = 5 754 + 17 ig = 5 5-754 17

Applying 17 = X 10-3 A Ruting KVL in loop 2, we get Rolat Relistig) + Roiz - Ez * 4812 + 17 litig) + 32 12 = 24 Itų + 97 iz = 24 Substituting 17 ist 97 5-7511 the value of by from (24817) - (9785) equation ③ /1EXIE)- / 8EX76) 11.02204 - 0.011022 A the value of is in equation ③ we get 5-76Y0011002) 17 0.24549 0.2455 {1+is) - 0.00022 A +. 0-2455 A 0.256 522 A 0.25652 A current through Ri, R2, R3 and R4 respectively (tial, is, ig and is. So 11 = (1,+iz) = 0.25652 A 19 = 0.2455 A Iz = 0.2455 A TH Dat = 0.011022 A A A are