h 14 V ЗА 60 Two sources of ideal emf and three resistors are connected as in the sketch. Two of the currents are known and the directions are indicated on the sketch. Apply Kirchoffs loop rule clockwise in the top loop. Twee bronne van ideale emk en drie weerstande word geskakel soos in die skets. Twee van die strome is bekend en die rigtings is op die skets aangedui. Pas Kirchoff se lusreël kloksgewys toe vir die boonste lus A. - 42 (3 A) - 14 V - 6 N (2 A) - E = 0 OB. 40 (1 A) - 14 V + 60 (2 A) - E = 0 O C. - 42 (1 A) + 14 V - 6 2 (2 A) + E = 0 O D. 4 2 (3 A) +14 V+6Q (2 A) + E = 0 OE. 4 2 (3 A) - 14 V + 6 2 (2 A) – E = 0
h 14 V ЗА 60 Two sources of ideal emf and three resistors are connected as in the sketch. Two of the currents are known and the directions are indicated on the sketch. Apply Kirchoffs loop rule clockwise in the top loop. Twee bronne van ideale emk en drie weerstande word geskakel soos in die skets. Twee van die strome is bekend en die rigtings is op die skets aangedui. Pas Kirchoff se lusreël kloksgewys toe vir die boonste lus A. - 42 (3 A) - 14 V - 6 N (2 A) - E = 0 OB. 40 (1 A) - 14 V + 60 (2 A) - E = 0 O C. - 42 (1 A) + 14 V - 6 2 (2 A) + E = 0 O D. 4 2 (3 A) +14 V+6Q (2 A) + E = 0 OE. 4 2 (3 A) - 14 V + 6 2 (2 A) – E = 0
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Chapter1: Units, Trigonometry. And Vectors
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Two sources of ideal emf and three resistors are connected as in the sketch. Two of the currents are known and the directions are indicated on the sketch. Apply Kirchoff's loop rule clockwise in the top loop.
Transcribed Image Text:b 14 V
3 A
60
Two sources of ideal emf and three resistors are connected as in the sketch. Two of the currents are known and the directions are indicated on the sketch. Apply Kirchoffs loop rule clockwise in the top loop.
Twee bronne van ideale emk en drie weerstande word geskakel soos in die skets. Twee van die strome is bekend en die rigtings is op die skets aangedui. Pas Kirchoff se lusreël kloksgewys toe vir die boonste lus.
O A. - 4 N (3 A) - 14 V - 6 2 (2 A) - E = 0
O B. 4 2 (1 A) - 14 V + 60 (2 A) – E = 0
O C. - 4 2 (1 A) + 14 V - 6 2 (2 A) + E = 0
O D. 4 2 (3 A) +14 V + 6 N (2 A) +E=0
Ο Ε.4Ω 3 Α) - 14 V +6Ω (2 Α) -E=0
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Introduction:
Kirchoff's second law, states that the amount of total energy in a closed remains conserved, and sum of the potential drop across resistor and emf of the batteries is equal to zero.
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