= E/R: I = 24/813 = 0.0295 A or 29.5 mA. A.4.2 Kirchhoff's Second Law When current flows through a resistor, there is always a difference in potential when measured across the resistor (i.e., between one end and the other of the resistor). The value in volts of the difference in potential varies with the current flowing through the resistor and the resistance. In fact, as mentioned, the value is the product of the resistance (in ohms) and current (in amperes). This is just the statement of Ohm's law, Eq. (A.1) (E = IR). In our previous discussion, it was called potential difference, voltage drop, or IR drop. This IR drop acts in the opposite direction to, or opposes, the emf which drives the current through the resistor. Figure A.7 is a model of a circuit with three resistors in series and a battery emf source. For reference, a voltmeter is shown measuring the voltage (the IR drop) across resistor R2. In such a closed circuit, the sum of the IR drops across the three resistors must be equal to the impressed emf. Let the IR drop across each resistor in Figure A.7 be represented by V1, V2, and V3, respectively. This, then, is how we state Kirchhoff's second law: E = V + V2 + V3, (A.12) A.4 RESISTANCES IN SERIES AND IN PARALLEL, AND KIRCHHOFF'S LAWS Figure A.7 A model circuit for three resistors in series. A voltmeter measures the IR drop across resistor R2. where E is the impressed emf or closed circuit battery voltage. In the case of Figure A.7, we can write E = IR1 + IR2 + IR3. (A.13) This equation can be restated as E - IR1-IR2-IR3 - 0. A.4.3 Hints on Solving dc Network Problems Make a network drawing, much like we have in Figures A.1 through A.7. Then assign letters to all unknown values. This should be followed by arrows showing the direction of current flow. With the practical application of Kirchhoff's laws, use correct algebraic signs. There should be one sign (+ or -) given to the electromotive force in the direction of the current flow, and the opposite sign is then given to the IR drops. Often we accept 