(13%) Problem 4: Consider the circuit diagram in the figure. Ez = 18V 0.5 02 R 250 2 1, R a pe 13 6.0Ω w RE 1.52 h elle
Posted: Tue Apr 26, 2022 7:12 pm
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(13%) Problem 4: Consider the circuit diagram in the figure. Ez = 18V 0.5 02 R 250 2 1, R a pe 13 6.0Ω w RE 1.52 h elle W 0.5Ω Es = 45V
A 50% Part (a) What is the equation which results when applying the loop rule to loop aedcba, in terms of the variables given in the figure? 0= & B 7 8 9 d 4 5 6 1 | 2 | 3 h 12 m 1 р R 11 + - 0 R 1 VO Submit Hints: Feedback: FA 50% Part (b) If the current through the middle part of the loop is 11 = 4.75 amps, what is the current through the top loop. I), in amps?