Question

25. In a balanced Wheatstone bridge, there is a. no load current b. maximum load current c. no source voltage d. equal current in the arms and the load 26. Refer to Figure 8 below. Assume R. = 2.0k12. The net Thevenin voltage of the bridge is R 1.0 k 2 R 2.0 k 2 +15 V R 3302 R 3.0 k2 R. Figure 8 a. 15 V b. 11.25 V c. 7.5 V d. 3.75 V 27. Refer to Figure 9 below. To balance the bridge, R. should be equal to: R. 1.0 k 2 R 2.0 k 2 +15 V = R 3302 R 3.0 k 2 R. Figure 9 a. 1.02 b. 2.0 k2 c. 4.0 k2 d. 6.0 k2

Answer 1

balanced Wheat stone Bridge 26 ) In There is j Ans: No load Current i ends of The ( voltages at {qual ) load aie 26 ) RI R2 = 2k2 Pi Elke Vs =15V 3 ke Rq = 2 kr -R3 Vith will remove find we O To 3 voltage load and finds open ckted volt across it. - loop -15VT (I, - [2) CIKt 3k) =0 =0 -150+ Car I2) (4K) a 15V 4K II -4K I 2 ZO Bk+ 1K) (12 -Il) + (2 kt 2k) I2 -6 = 0 -1, (46) + 86 12 @& b) Il = 7.5mA c By solving I 22 3.75mA

Now - Vnt (12-1) Rit I2 R2 =0 - Vino Tik(3.75 mt - 7,5mlJet 3.75 mA (26) =0 - UTO + (-3. 75 )Vt 7.5 V =0 Uth t 3.750 =0 3 75V Uth= ckt Bridge 27) To bila lance given RI R2 - 25 R4 R, R4 R3 R2 =) (IKR) (R4) (3km) (akas 6 kr - R4