Question

-First, determine the equivalent resistance of this car's electrical circuit.

-Second, what is the current in each branch if the applied voltage is a 12 V car battery? (Be sure to track the current to each place it splits!)

2Ω 4Ω A 3Ω 8Ω 6Ω 6Ω B

Answer 1

2.2 ча 22 Ad 32 312 ww ER2 812 =) Ri. GA & 6.2 BO (B) (A2 2.12 212 AO Ry € Rz im R2 Bo- (D) (C) 62 & 62 By figure 'A' and us &82 are in Parellel and Series Combination shes pedively so, ♡ = 322 6x6 6+6 = 4 +8 = 122 3 R2 → By Figure 'B' 328 R, is in senies I R₂ 3+ R, 3+3 = 6.2 -) in parellel Figure 'C' Rp & R2 are Ru = R₂R R2 R3 6812 6+12 = 42 By figure 'D' RAB at Ru 602

2.2 22 ua Em Ao P y Iz 32 I I, I Il y I, 832 R2 8.2 =) I3 I3 R 0.) mus & G2 D Bo BG Q (B) (A2 22 V P 252 D I Ru R3 ww R2 Bo- (D) (c)

→ By Figure'b'value of current 'T' is 1 I = V 12 6 = 2A RAB 2X41 = 8V so, Current through 202 R2 I = 2A - By Figure (D) Voltage accross Ru I Van Iru = =) By Figure (B] Value of 'I'EI' are ? = Voo = 0.666 A Therefore, By Figure A' current through Ii 8 12 R2 ☺ = Irun for Rus => = I 0.666 A Ros =) Trea I, 0.666 A I2 = Voo = 8 3+3 = 1,333 A 3+RI There fore, current through 312 is IRBR I 2 1.333 A Now, By figure 'A' Iz will divided equally in Ca So, current through 6127 Branch IRER = I, - Iz 0.666 A 2