29.1. Electromotive Force
2. Electric Generators: Electric generators convert mechanical energy into electric energy. The principle of operation of electric generators will be discussed in Chapter 31.
3. Fuel cells: A mixture of chemicals are combined in the fuel cell. The chemical reactions that occur are
4. Solar Cell: A solar cell converts the energy of the sunlight directly into electric energy. The emf of a silicon solar cell is 0.6 V. However, the amount of current that can be extracted is rather small.
27.2. Single-loop currentsA source with a time-independent emf is represented by the symbol shown in Figure 29.2. The long line in Figure 29.2 represents the positive terminal of the source, while the short line represents the negative terminal.
In this sum, the emf of a source is reckoned as positive if the current flows through the source in the forward direction (from the negative terminal to the positive terminal) and negative if it flows in the backward direction. (Note: the direction of the current indicates the direction of the positive charge carriers).
29.3. Multi-loop circuits.The procedure used to calculate the current flowing through a complicated circuit (with several resistors and emfs) is called the loop method. This procedure can be summarized as follows (see Figure 29.5 and Problem 29.17):
2. Label the current in the loops I1, I2, I3, ....., and arbitrarily assign a direction to each of the currents. A useful policy can be to define the direction of the current in a loop as going from the positive terminal of the emf to the negative terminal of the emf (if an emf is present in the loop). This policy defines the direction of the current in loop 1, loop 2 and in loop 3 (see Figure 29.6).
3. Apply Kirchhoff's rule to each loop.
The sum of all currents entering a branch point of a circuit (where three or more wires merge) must be equal to the sum of the currents leaving the branch point.
The branch method involves the following steps (and is illustrated by discussing its application to problem 29.17):
2. Label the currents in each branch I1, I2, I3, ....., and arbitrarily assign a direction to each of these currents. The currents in the five branches of circuit 29.5 are indicated in Figure 29.7.
3. Apply Kirchhoff's second law to each loop.
4. Apply Kirchhoff's first law to each branch point.
For loop 1 of the circuit shown in Figure 29.7 Kirchhoff's law dictates that
The current I1 can be obtained from eq.(29.14):
Example: Problem 29.10Consider the circuit shown in Figure 29.8. Given that [epsilon]1 = 6.0 V, [epsilon]2 = 10.0 V, and R1 = 2.0 [Omega], what must be the value of the resistance R2 if the current through this resistance is to be 2.0 A ?
29.4. Energy in circuitsTo keep a current flowing in a circuit, work must be done on the circulating charges. If a charge dq passes through a battery with emf [epsilon], the work done dW will be equal to
The moving charges dissipate some of their energy when passing through resistors. Suppose the potential drop across a resistor is [Delta]V. For an electron moving through the resistor the loss of potential energy is equal to e . [Delta]V. The energy lost is converted into heat, and the rate at which energy is dissipated is equal to
Example: Example 8A high-voltage transmission line that connects a city to a power plant consists of a pair of copper cables, each with a resistance of 4 [Omega]. The current flows to the city along one cable, and back along the other.
a) The transmission line delivers to the city 1.7 x 105 kW of power at 2.3 x 105 V. What is the current in the transmission line ? How much power is lost as Joule heat in the transmission line ?
b) If the transmission line delivers the same 1.7 x 105 kW of power at 110 V, how much power would be lost in Joule heat ? Is it more efficient to transmit power at high voltage or at low voltage ?
a) The power delivered to the city, Pdelivered, is equal to 1.7 x 105 kW. The voltage delivered, [Delta]Vdelivered, is equal to 2.3 x 105 V. The current through the cables can determined by applying eq.(29.32):
b) If the voltage delivered to the city is 110 V, than current through the transmission cables must be equal to
Example: Problem 29.24.A 12-V battery of internal resistance Ri = 0.20 [Omega] is being charged by an external source of emf delivering 6.0 A.
a) What must be the minimum emf of the external source ?
b) What is the rate at which heat is developed in the internal resistance of the battery ?
a) The circuit describing this problem is shown in Figure 29.9. In Figure 29.9 [epsilon]1 is the emf of the battery being recharged, and [epsilon]2 is the emf of the recharger. Applying Kirchhoff's second rule to the single loop circuit we obtain the following relation between the charging current and the emfs:
Example: Problem 29.32Suppose that a 12-V battery has an internal resistance of 0.40 [Omega].
a) If this battery delivers a steady current of 1.0 A into an external circuit until it is completely discharged, what fraction of the initial stored energy is wasted in the internal resistance ?
b) What if the battery delivers a current of 10.0 A ? Is it more efficient to use the battery at low current or at high current ?
a) Suppose the battery as an emf [epsilon]b and delivers a current I. The internal resistance of the battery is equal to Ri. The voltage drop across the internal resistance is equal to I Ri. The external voltage of the battery if thus equal to ([epsilon]b - I Ri). The power delivered by the battery to the external circuit is therefore equal to
29.5. The RC circuitThe current through the circuits discussed so far have been time-independent, as long as the emf of the source is time-independent. The currents in these circuits can be determined by applying Kirchhoff's first and/or second rule.
A simple circuit in which the current is time dependent is the RC circuit which consists of a resistor R connected in series with a capacitor C (see Figure 29.10). Applying Kirchhoff's second rule to the current loop I gives