Chapter 20. -SHIPBOARD ELECTRICAL SYSTEMS 



as the power supply, such as a battery. The 

 voltage drop is usually regarded as the load, 

 such as a resistor. The voltage drop may be 

 distributed across a number of resistive ele- 

 ments, such as a string of lamps or several 

 resistors. However, according to Kirchhoff's 

 law, the sum of their individual voltage drops 

 must always equal the voltage rise supplied by 

 the power source. 



The statement of Kirchhoff's law can be 

 translated into an equation, from which many 

 unknown circuit factors may be determined. 

 (See fig. 20-6.) Note that the source voltage Eg 

 is equal to the sum of the three load voltages 

 Ej, E2, and E3. In equation form, 



E = E + E + E 

 s 1 2 3 



The following procedure may be used to solve 

 problems applicable to figure 20-6: 



1. Note the polarity of the source emf (Eg) 

 and indicate the electron flow around the circuit. 

 Electron flow is out from the negative terminal 

 of the source, through the load, and back to the 

 positive terminal of the source. In the example 

 being considered, the arrows indicate electron 

 flow in a clockwise direction around the circuit. 



2. To apply Kirchhoff's law it is necessary 

 to establish a voltage equation. The equation is 

 developed by tracing around the circuit and not- 

 ing the voltage absorbed (that is, the voltage 

 drop) across each part of the circuit, and ex- 

 pressing the sum of these voltages according to 

 the voltage law. It is important that the trace 

 be made around a closed circuit, and that it 

 encircle the circuit only once. Thus, a point is 

 arbitrarily selected at which to start the trace. 

 The trace is then made and, upon completion, 



the terminal point coincides with the starting 

 point. 



3. Sources of emf are preceded by a plus 

 sign if, in tracing through the source, the first 

 terminal encountered is positive; if the first 

 terminal is negative, the emf is preceded by a 

 minus sign. 



57 Voltage drops along wires and across 

 resistors (loads) are preceded by a minus sign 

 if the trace is in the assumed direction of elec- 

 tron flow; if in the opposite direction, the sign 

 is plus. 



5. If the assumed direction of electron flow 

 is incorrect, the error is indicated by a minus 

 sign preceding the current, as obtained in solv- 

 ing for circuit current. The magnitude of the 

 current is not affected. 



The preceding rules may be applied to the 

 example of figure 20-6 as follows: 



1. The left terminal of the battery is nega- 

 tive, the right terminal is positive, and electron 

 flow is clockwise around the circuit. 



2. The trace may arbitrarily be started at 

 the positive terminal of the source and con- 

 tinued clockwise through the source to its nega- 

 tive terminal. From this point the trace is con- 

 tinued around the circuit to a, b, £, d, and back 

 to the positive terminal, thus completing the 

 trace once around the entire closed circuit. 



3. The first term of the voltage equation is 

 +Eg. 



4. The second, third, and fourth terms are, 

 respectively, -Ej, -E2, -E3. Their algebraic 

 sum is equated to zero, as follows: 



^2-^3 







13.15 

 Figure 20-6.— Series circuit for demonstrating 

 Kirchhoff's law of voltages. 



^s 1 



Transposing the voltage equation and solving 

 for Eg, 



^s = El + E2 + Eg 



Since E = IR, from Ohm's law, the voltage 

 drop across each resistor may be expressed in 

 terms of the current and resistance of the indi- 

 vidual resistor, as follows: 



Eg = IRj + IR2 + IR3 



where Rj, R2, and R3 are the resistances of 

 resistors Rl, R2, and R3, respectively. Eg is 

 the source voltage and I is the circuit current. 



E may be expressed in terms of the circuit 

 current and total resistance as IR^. Substituting 

 IR^ for Eg, the voltage equation becomes 



IR. = IRj + IRg + IR3 



497 



