With the loop theorem, it is straightforward to solve for the current through any single loop circuit. Let's consider a circuit with three resistors r one, r two, and r three. Let's ignore the internal resistance this time. Apply the loop theorem. We will begin at point A. Proceed in the direction of the current flow. The potential increases by epsilon across the e m f. It decreases by I R one across the first resistor, by I R two across the second resistor and by I R three across the third resistor. Once again, the equation can be solved for the current, I. When the same current flows through multiple circuit elements, we say that they are in series. The three resistors shown here are in series. When resistors are in series, as in this case, the voltages across each resistor are proportional to the current. Let's consider the general case of some number N of resistors. The voltage across each resistor is the current times the resistance of that resistor. The current can be factored out on the right hand side. The combined effect of all the resistors is the same as one equivalent resistance R sub e q, which is equal to the sum of the individual resistances. As we will see later in this unit, for more complex circuits, the effect of a network of resistors can be reduced to an equivalent resistance. For resistors in series, the equivalent resistance is equal to the sum of the individual resistances. A circuit has two resistors and a battery. The internal resistance of the battery is 0.2 ohms, and the resistors have resistances of 0.6 ohms and 0.4 ohms. The e m f is 5 Volts. What is the terminal voltage? ... Make sure not to confuse current and voltage ... The terminal voltage is less than the emf. ... The e m f is 5 Volts. What is the terminal voltage? ... To make a point about using the loop theorem, let's return to a simple circuit that includes only an e m f and a resistor. In this simple circuit, the direction of current flow is obvious. Sometimes, in more complex currents, it is not so obvious. Let's solve for the current in this circuit under the assumption that the current is flowing as shown - in the wrong direction. When we integrate around the circuit in the same way as before, the jump in voltage across the e m f remains the same. Because we are traveling against the assumed current direction, the voltage increases across the resistor. Next we'll solve this for the current. The result is the same as before, except the sign is now negative. This indicates that the current flow is opposite the assumed direction. Applying the loop theorem gave the same result, even though the assumed current direction was wrong. In general, when we solve circuit problems, we can simply choose a direction for the current. A negative result indicates the current is actually flowing in the opposite direction. The circuit shown includes two batteries, each with an internal resistance, and an additional resistor, R one. Note that the two e m f's are in opposite directions. We will apply the loop theorem to the circuit shown, beginning by moving in the direction of I across the first battery in the direction of its e m f. The voltage will increase by e one and decrease by I times small r one. The circuit path goes through the second battery. Because we are traveling in the direction opposite to the e m f of this battery, the voltage decreases by e two and by I times small r two. Finally, we complete the circuit by crossing resistor R and returning to the starting point. Solve this equation for the current. If epsilon one is greater then epsilon two, the current in the second battery will flow against its e m f. This indicates that energy is being added to this battery; that is, it is being recharged. Using this type of arrangement won't recharge most real batteries, however. In fact, the battery may explode. In this circuit, epsilon one is 5 volts, epsilon two is 2 volts, small r one and small r two are both 0.4 ohms, and capital R is 2 ohms. What current is flowing though the circuit? ... Both emf's are in the same direction. ... Both emf's are in the same direction. ... Make sure you have included all the resistances in your calculation. ... In the circuit shown, a current of 2.5 amperes is flowing through the circuit. What is the voltage across the 2 ohm resistor? Use Ohm's Law. ... What is the net power being generated by the first battery with an e m f epsilon one? The current is 2.5 amperes. ... Do not include the power dissipated by the internal resistance. ... The answer should be in Watts. ... In this circuit, we have a battery with an internal resistance. For what value of resistance, R one, is the power dissipation in this resistor the greatest? The power dissipated in the resistor R one is the current squared times r one. The internal and external resistances are in series, and we can write down an expression for the current. Substitute this representation into the expression for the power. Let's plot the power as a function of R one. The axes have been put in terms of e and r. The important thing to note is that at first the power increases as R one increases, but then it decreases. Let's calculate the maximum value of the power. How do we find the location of the maximum value of this function? We want the location of the maximum. ... The derivative is zero at the maximum. ... To find the maximum, take the derivative of the power as a function of R1. The result is as shown. This condition is satisfied when R one is equal to the internal resistance, small r. Using this result in the equation for the power shows that the maximum power is one fourth epsilon squared divided by the internal resistance.

Copyright 2006 The Regents of the University of California and Monterey Institute for Technology and Education