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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

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

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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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.

b 14 V 4Ω 3 A E 2A 62 R 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. 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 2 (3 A) - 14 V - 6 2 (2 A) - E = 0 O B. 40 (1 A) - 14 V + 62 (2 A) - E = 0 O C. - 4 2 (1 A) + 14 V - 6 Q (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 Reset Selection
Transcribed Image Text:b 14 V 4Ω 3 A E 2A 62 R 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. 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 2 (3 A) - 14 V - 6 2 (2 A) - E = 0 O B. 40 (1 A) - 14 V + 62 (2 A) - E = 0 O C. - 4 2 (1 A) + 14 V - 6 Q (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 Reset Selection
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The given figure is, 

Physics homework question answer, step 1, image 1

Clearly it is visible that the circuit contains two EMF sources and three resistance. 

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