Question

16. What is the magnitude of the current in the 2012 resistor? (4pts) 100 w 10 V 17. A rather adventurous scientist flying in a hot air balloon above the ground creates artificial lightning using his specially designed machine. Assume each lightning bolt strikes the ground vertically and acts like a current in a long straight wire. Each one induces an emf m a large ning (or hoop) standing vertically (upright) on the ground. What emf is induced in the ring (diameter OSm itus A5.0m.ficom. 2.00x LUPA lightning bold The current is to zero. in 22.Busants

Answer 1

Ta arablen can be solved by using superposition the arena in the given ciraut, Tor According to superposition theorem we cessume one battery at a time and other is short. Then we fond current through 200 for bath case. Now add the current which we found in bath case. The final value is the actual value of current which is passes through 20. Case io 100 so lov un / 2012 Y o-i) T IL بان ۱۵ Now, in mash I 10 v = (isxlov ) + ,x(2012) – 0 in mush I lov = (x102) + cio-ijx100 - ② 100 = 1010 + 1060 - 10i, 10 = dois - 101, į, a to (10 + loi) - Put this value in equation o lov = 10 +10 in) x10 + 20, 10 = {1+1 5+5 i, +208,

10 = 5+51, +20 5 = 251, [ = 15 Amp] = 1/2 Amp Now, case I 15V GI is lou Now in mash I 150=20 iz + 10 i. - In mash I 150 = 10 cis-in + 1080 -6 15= 1080 - 10i, tloi! 15= 2010 - 10 i2 1. = 110 (15 +10 (2) - ③ Put this value in equation ④ 15 = 2012 +lo x 115 +10 in) 15=2012 + 15 + 5i2 25i2 = 15 - rç = 30;15 25i2 = 15 in 1926 - 2351

Now according to seefperybosition theorena, the actual current through 2012 is, i= ( 3 + 3 ) Amp iz (2 + 3 12 ) Amp i = 2 Amp i = 1/2 Amp Ti = 0.5 Amp