(40)Problem 1 (SeeFigs. 1)Useassigned coordmate system.Answer questions a e. 24.
ID: 1430691 • Letter: #
Question
(40)Problem 1 (SeeFigs. 1)Useassigned coordmate system.Answer questions a e. 24.0 V 12.0 kn 1 0.0F 6.00 020 R3.00 R-15.0 kn 3,00 0 Figurela: Batteries are ideal Use signed currents and loops (zad2) to find answers Figurelb Use Fig. la to answer a-d. a. Apply Junction Rule to nodeafor the given current assignments b. Apply Kirchhoff s Loop Rule to loops zand zas assigned| c. Calculate by following the path of current I; calculate the values of currents/, and I,. Is current I,in the direction imdicated in Fig. la? Yes, No] Explain Circle Oue d. Calculate iveR-8.0 Use Fig. 1b to answere-f. e. Withswitch closed the capacitor becomes fully charge and steady state is achieved, fnd the value ofthe currents through all three resistors. f. Withthe maximum chargeof 500"available on the capacitor the switch is opened. The capacitor immediately begins tolcharge, discharge, explode·Calculate the tme constant related to this condition Circle Oue andwrite the specific equation govening this current.Explanation / Answer
a) I3 = I1 + I2
b) For x loop
E = 3I2 + 8I3
For y loop
24 = 18 - 3I2
c) Current I2 = 2 A
Current I3 = 1 A
=> Vab = 3 * 2 = 6V
Also , Is current I2 in direction indicated ----- NO
d) Here, E = 8*1 -3*2
= 2 V
e) current through 12 kohm = 9/27
= 0.333 mA
current through 15 kohm = 9/27
= 0.333 mA
current through 3 kohm = 0 mA
f) time constant = 18 * 103 * 10 * 10-6
= 0.18 sec
The current immediately begin to discharge .
Here, peak voltage = 50/10
= 5 V
peak current = 5/18
= 0.277 mA
=> current equation = 0.277 * 10-3 (e-t/0.18)
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.