Consider the circuit diagram in the figure. 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? Part (b) If the current through the middle part of the loop is I1 = 4.75 amps, what is the current through the top loop, I2, in amps?

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Consider the circuit diagram in the figure. 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? Part (b) If the current through the middle part of the loop is I1 = 4.75 amps, what is the current through the top loop, I2, in amps?
E= 18 V
%3D
in
0.5Q
2.5Q
:-
R,
a
6.0 2
1.5 2
6.
12
0.52
E2 = 45 V
%3D
Transcribed Image Text:E= 18 V %3D in 0.5Q 2.5Q :- R, a 6.0 2 1.5 2 6. 12 0.52 E2 = 45 V %3D
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