Question

6. [8/30 Points) DETAILS PREVIOUS ANSWERS SERPSE 10 27.3.P.017.MI. The circuit shown in the figure below is connected for 2.90 min. (Assume R = 6.20 , R2 = 2.00 n, and V = 16.0 v.) 5.00 3.00 22 R2 1.0012 R + -4.00 V V direction down (a) Determine the current in each branch of the circuit. branch magnitude (A) left branch X Your response differs from the correct answer by more than 10%. Double check your calculations. middle branch right branch down up (b) Find the energy delivered by each battery. 4.00 V battery 16.0V battery ] (c) Find the energy delivered to each resistor. 6.2022 resistor 5.00 resistor 1.00 A resistor 3.00 n resistor 2.00 A resistor (d) Identify the type of energy storage transformation that occurs in the operation of the circuit. Score: 2 out of 2 Comment: (e) Find the total amount of energy transformed into internal energy in the resistors.

Answer 1

t=2.9 min R,& in all) Using R2=22 Iz ing Iv- 16v bür Thu KCL Ist I = Is Using KVL in lift loop 61, +6.2 Iz = 4 Usning KVL in right loop (2+3) I 2 - 6 I, t4-16=0 -6I, + 512= - (2) 12 -) - I 2 I3 -Ⓡ) Solving aq 112, (2), (3), one get I = -0.55LA =1.735A 1:181 A a left branch 1.181A Midde brouch - 10.SSHA Right bearby [1.735 A (b) Energy by 4V - 4X0.55412-9x60 - 385.584J Fenergy by 16v 16+ 1.735+ 2.9460 - (4830.24/5

El 6.25 resistor = (181)"+6:2+2.9x60=1504.7J sr resistor = (0.554)²x5+2.9460 =267 J in resistor - (6.554)²x1x2.9.60 = 53.45 or resista = 1.735)²+ 3+2.4460 = 1571-35 ar resistor - (1.735)²+2+2.9x60=1047.50 (d) Energy is wasted in the form of heat