(8%) Problem 12: Consider the circuit in the figure, with the current directions
ID: 1864684 • Letter: #
Question
(8%) Problem 12: Consider the circuit in the figure, with the current directions defined as shown. There are four resistors in this circuit (Ri, R2, R3, and R4) and four batteries with emfs = 19 V?-4 S ? = 12 V and 4-22.5 V each with internal resistance given by ri as marked in the figure 0.52 Randomized Variables 19 V -4.5 V 63-12 V 4 22.5 V 0.25 ? RA 15? 0.75 ? ©theexpertta.com 33% Part (a) Calculate the current ? in amps .ilà 33% Part (b) Calculate the current ?2 in amps 3300 Part (c) Calculate the current /3 in ampsExplanation / Answer
b) c) Applying KCL at node point ‘a’,
I1=I2+I3 -----------(1)
Applying KVL in loop1,
E1-I1r1-I1R1-I3R2+E2-I3r2-I1R4 = 0
19+4.5 – I1(0.5+20+15)-I3(6+0.25)=0
23.5 - 35.5 *I1- 6.25*I3 = 0 -------------(2)
Plug (1) in (2)
23.5 - 35.5 *(I2+I3)- 6.25*I3 = 0
23.5 -35.5*I2-35.5*I3 – 6.25*I3 = 0
23.5 -35.5*I2-41.75*I3 = 0 ----------------(3)
Applying KVL in loop2,
-I2R3+E3-I2r3-I2r4-E4+I3r2-E2+I3R2 = 0
-I2(8.0+0.75+0.50)+I3(6.0+0.25)+12-22.5-4.5 = 0
-9.25*I2+6.25*I3-15 = 0 -------------(4)
Solving (3) and (4) as simultaneous eqns,
I2= - 0.788A and I3= 1.23A ……………negative sign indicates actual direction of the current is reverse as shown the figure
a)Plugging these values in (1),
I1= -0.788+1.23 = 0.442A
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