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www.GOALias.blogspot.comPhysicsExample 3.5 A network of resistors is connected to a 16 V batterywith internal resistance of 1Ω, as shown in Fig. 3.19: (a) Computethe equivalent resistance of the network. (b) Obtain the current ineach resistor. (c) Obtain the voltage drops V AB, V BCand V CD.112EXAMPLE 3.5FIGURE 3.19Solution(a) The network is a simple series and parallel combination ofresistors. First the two 4Ω resistors in parallel are equivalent to aresistor = [(4 × 4)/(4 + 4)] Ω = 2 Ω.In the same way, the 12 Ω and 6 Ω resistors in parallel areequivalent to a resistor of[(12 × 6)/(12 + 6)] Ω = 4 Ω.The equivalent resistance R of the network is obtained bycombining these resistors (2 Ω and 4 Ω) with 1 Ω in series,that is,R = 2 Ω + 4 Ω + 1 Ω = 7 Ω.(b) The total current I in the circuit isε 16VI = = = 2AR+ r (7 + 1) ΩConsider the resistors between A and B. If I 1is the current in oneof the 4 Ω resistors and I 2the current in the other,I 1× 4 = I 2× 4that is, I 1= I 2, which is otherwise obvious from the symmetry ofthe two arms. But I 1+ I 2= I = 2 A. Thus,I 1= I 2= 1 Athat is, current in each 4 Ω resistor is 1 A. Current in 1 Ω resistorbetween B and C would be 2 A.Now, consider the resistances between C and D. If I 3is the currentin the 12 Ω resistor, and I 4in the 6 Ω resistor,I 3× 12 = I 4× 6, i.e., I 4= 2I 3But, I 3+ I 4= I = 2 A⎛Thus, I 3= 2 ⎞⎜⎝⎟3⎠ A, I = ⎛4⎞4⎜⎝⎟3⎠ Athat is, the current in the 12 Ω resistor is (2/3) A, while the currentin the 6 Ω resistor is (4/3) A.(c) The voltage drop across AB isV AB= I 1× 4 = 1 A × 4 Ω = 4 V,This can also be obtained by multiplying the total current betweenA and B by the equivalent resistance between A and B, that is,