Question

7. In the figure below, what is the net resistance of the circuit connected to the battery? Each resistance in the circuit is equal to 16.00 k92. kΩ 8. Find the current in the R2 resistor in the drawing (V1 = 11.0 V, V2 = 22.0 V, V3 = 14.0 V, R1 = 1.80 92, R2 = 2.40 2, and R3 = 3.00 92).

Answer 1

wML R- 16 k = 16000 cincuit can be simplified by successive educhon me thod Rasiston ,2,3,4 Gnd are connected poorallel, hene theire equivalant sesistane is and the cincuit is, R and R are in series, and thein equivalont Nalus R+ 5 R R// 5R 5R RW R Rx 5R R 5R2 RT5R 5R 5R R 5R and Rane in series RS 5R R=148 1 R

14R (4R R R 14 R X 148 R + 4 R2 t2&R+9R 14 R 148 37 37R 148 37 R+ I4R 37 37 R+14R 5IR 37 Roy= 37 51x 16000 37 22054.05 22.05405 k Rey 22 k

R I c R1-8 M R-2-42 Rs-3 22V V UN 1IN 14N=V3 A V=14V AL junction C, -T+I Using KVL for CDEFC I, R.-V- +T, RO R22 2.4I-22-14 31,= Using KV L for BcFAB - V O I,R, R+ 1-8.1,-313 +4-11 6 1-8I-3I +3 o a.4I-I-36 0 2.4 CI+1,)-3I-36-0 using 2.4T3- 3T, -36-0 2.4 I, 2.41,-0.6-36 0 0-61- -12 6 0.6 4 +) +) I3+ 0.6. T,-4 = O 47 A I (-6.85) 0.85

1-81,-3x(-4.7) +3-6 Now 3- 1-81-14.l+3=0 1.&I 11-1 I 11.1 6.17 A I 1+I Now G.IT-4.7 1-47 A Curent though R is T= 1.47A

Answer 2

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