D.C. Circuits 55 Substituting this value for I 1 into equation [3] gives: 1111111.( ..).....550 95254272242750 07222IIIhence, 3320 03662and A I.Ans Note: The minus sign in the answer for I 2 indicates that this current is actually fl owing in the opposite direction to that marked in Fig. 2.22 . This means that battery E 1 is both supplying current to the 5 resistor and charging battery E 2 . Current through 5 resistor ampso cuII1120 950 0366.(.)rrrent through 5 resistor A 0 950 03660 9 4...11Ans (b) To obtain the p.d. across the 5 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 emf E 2 , because E 2 is being charged. A third alternative is to multiply R 3 by the current fl owing through it. All three methods will be shown here, and, provided that the same answer is obtained each time, the correctness of the answers obtained in part (a) will be confi rmed. VERVBEBE11 1111I voltso, V 60 955642654 574(.. )..Ans OR: voltso, VERVBEBE22 24 50 036624 50 07324 573I.( .)...VV Ans OR: VRVBEBE()..II112309 45457 voltso, V Ans The very small diff erences between these three answers is due simplyto rounding errors, and so the answers to part (a) are verifi ed ascorrect. 2.8 The Wheatstone Bridge Network This is a network of interconnected resistors or other components, depending on the application. Although the circuit contains only one source of emf, it requires the application of a network theorem such as the Kirchhoff ’ s method for its solution. A typical network, suitably labelled and with current ﬂ ows identiﬁ ed is shown in Fig. 2.23 .