Fundamental Electrical and Electronic Principles, Third Edition

D.C. Circuits 55Substituting this value for I 1 into equation [3] gives:1. 5( 1. 50. 951)2I21. 51.4272I2hence, 2I21. 4271. 50.0732and I 0.0366 A Ans2Note: The minus sign in the answer for I 2 indicates that this current isactually flowing in the opposite direction to that marked in Fig. 2.22 . Thismeans that battery E 1 is both supplying current to the 5 Ω resistor andcharging battery E 2 .Current through 5 Ω resistor I1 I2 amp 0. 951( 0. 0366)so current through 5 Ω resistor 0. 9510. 0366 0.914A Ans(b)To obtain the p.d. across the 5 Ω resistor we can either subtract the p.d.(voltage drop) across R 1 from the emf E 1 or add the p.d. across R 2 to emfE 2 , because E 2 is being charged. A third alternative is to multiply R 3 by thecurrent flowing through it. All three methods will be shown here, and,provided that the same answer is obtained each time, the correctness ofthe answers obtained in part (a) will be confirmed.so, VVBE E1I 1R1 volt 6( 0. 9511. 5)61.4265BE 4. 574 V AnsOR:so, VVBE E2I 2R2 volt 4. 5( 0. 03662)4. 50.0732BE 4. 573 V AnsOR:VBE( I1 I2) R3 volt 09 . 145so, V 457 . V AnsBEThe very small differences between these three answers is due simplyto rounding errors, and so the answers to part (a) are verified ascorrect.2.8 The Wheatstone Bridge NetworkThis is a network of interconnected resistors or other components,depending on the application. Although the circuit contains only onesource of emf, it requires the application of a network theorem suchas the Kirchhoff ’ s method for its solution. A typical network, suitablylabelled and with current flows identified is shown in Fig. 2.23 .