Principles of naval engineering - Historic Naval Ships Association

Chapter 20. -SHIPBOARD ELECTRICAL SYSTEMSas the power supply, such as a battery. Thevoltage drop is usually regarded as the load,such as a resistor. The voltage drop may bedistributed across a number of resistive elements,such as a string of lamps or severalresistors. However, according to Kirchhoff'slaw, the sum of their individual voltage dropsmust always equal the voltage rise supplied bythe power source.The statement of Kirchhoff's law can betranslated into an equation, from which manyunknown circuit factors may be determined.(See fig. 20-6.) Note that the source voltage Egis equal to the sum of the three load voltagesEj, E2, and E3. In equation form,E = E + E + Es 1 2 3The following procedure may be used to solveproblems 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 terminalof the source, through the load, and back to thepositive terminal of the source. In the examplebeing considered, the arrows indicate electronflow in a clockwise direction around the circuit.2. To apply Kirchhoff's law it is necessaryto establish a voltage equation. The equation isdeveloped by tracing around the circuit and notingthe voltage absorbed (that is, the voltagedrop) across each part of the circuit, and expressingthe sum of these voltages according tothe voltage law. It is important that the tracebe made around a closed circuit, and that itencircle the circuit only once. Thus, a point isarbitrarily selected at which to start the trace.The trace is then made and, upon completion,13.15Figure 20-6.— Series circuit for demonstratingKirchhoff's law of voltages.the terminal point coincides with the startingpoint.3. Sources of emf are preceded by a plussign if, in tracing through the source, the firstterminal encountered is positive; if the firstterminal is negative, the emf is preceded by aminus sign.57 Voltage drops along wires and acrossresistors (loads) are preceded by a minus signif the trace is in the assumed direction of electronflow; if in the opposite direction, the signis plus.5. If the assumed direction of electron flowis incorrect, the error is indicated by a minussign preceding the current, as obtained in solvingfor circuit current. The magnitude of thecurrent is not affected.The preceding rules may be applied to theexample of figure 20-6 as follows:1. The left terminal of the battery is negative,the right terminal is positive, and electronflow is clockwise around the circuit.2. The trace may arbitrarily be started atthe positive terminal of the source and continuedclockwise through the source to its negativeterminal. From this point the trace is continuedaround the circuit to a, b, £, d, and backto the positive terminal, thus completing thetrace 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 algebraicsum is equated to zero, as follows:^2-^3^s 1Transposing the voltage equation and solvingfor Eg,^s = El+ E2 + EgSince E = IR, from Ohm's law, the voltagedrop across each resistor may be expressed interms of the current and resistance of the individualresistor, as follows:Eg = IRj + IR2 + IR3where Rj, R2, and R3 are the resistances ofresistors Rl, R2, and R3, respectively. Eg isthe source voltage and I is the circuit current.E may be expressed in terms of the circuitcurrent and total resistance as IR^. SubstitutingIR^ for Eg, the voltage equation becomesIR.= IRj + IRg + IR3497