Fundamental Electrical and Electronic Principles, Third Edition

Fundamentals 15then current I will flow as shown. The value of this current will be5.71 A (the method of calculating this current will be dealt withearly in the next chapter). This current will cause a p.d. across r andalso a p.d. across R . These calculations and the consequences for thecomplete circuit now follow:p.d. across r Ir volt (Ohm’s law applied)571 . 01. 0.571Vp.d. across RIR volt571 . 2 11.42 VNote: 0.571 11.42 11.991 V but this figure should be 12 V. Thevery small difference is simply due to ‘ rounding ’ the figures obtainedfrom the calculator.IAr0.1 ΩE12 VVR2 ΩFig. 1.10BThe p.d. across R is the battery terminal p.d. V. Thus it may be seenthat when a source is supplying current, the terminal p.d. will alwaysbe less than its emf. To emphasise this point let us assume that theexternal resistor is changed to one of 1.5 Ω resistance. The current nowdrawn from the battery will be 7.5 A. Hence:p.d. across r 75 . 01 . 075. Vand p.d. across R 75 . 15 . 1125. VNote that 11.25 0.75 12 V (rounding error not involved). Hence,the battery terminal p.d. has fallen still further as the current drawn hasincreased. This example brings out the following points.1 Assuming that the battery ’ s charge is maintained, then its emfremains constant. But its terminal p.d. varies as the current drawn isvaried, such thatV E Ir volt (1.6)2 Rather than having to write the words ‘ p.d. across R ’ it is moreconvenient to write this as V AB , which translated, means thepotential difference between points A and B.