# 29.1. Electromotive Force

**Figure 29.1. Electron in electronic circuit.**

**. The emf of a source is defined as the amount of electric energy delivered by the source per Coulomb of positive charge as charge passes through the source from the low-potential terminal to the high-potential terminal. A steady current will flow through the circuit in Figure 29.1 if the increase of potential energy of the electrons in the source is equal to the change in the potential energy of the electrons along their path through the external circuit. The unit of emf is the Volt (V) and usually the emf is simply called the voltage of the source. The symbol for emf is [epsilon]. The most important sources of emf are: 1.**

__electromotive force (emf)__**: Batteries convert chemical energy into electric energy. In a lead-acid battery the following reactions take place when the battery delivers a current:**

__Batteries___{ }

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2.

**: Electric generators convert mechanical energy into electric energy. The principle of operation of electric generators will be discussed in Chapter 31.**

__Electric Generators__3.

**: A mixture of chemicals are combined in the fuel cell. The chemical reactions that occur are**

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4.

**: 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.**

__Solar Cell__## 27.2. Single-loop currents

**A 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.**

**Figure 29.2. Schematic symbol of source of emf.**

**, which states: Around a closed loop in a circuit, the sum of all the emfs and all the potential changes across resistors and other circuit elements must equal zero.**

__Kirchhoff's rule__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).

**Figure 29.3. Simple single-loop circuit.**

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**Figure 29.4. Real source consisting of internal resistance R**

_{i}and ideal emf [epsilon]._{i}of the source into account. The internal resistance R

_{i}of the source can be regarded as connected in series to an ideal emf (see Figure 29.4). If a current I is flowing though the source, the emf across the external terminals of the source is equal to [epsilon] - I

^{.}R

_{i}. The external emf therefore depends on the current delivered by the source.

## 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**

**. This procedure can be summarized as follows (see Figure 29.5 and Problem 29.17):**

__loop method__**Figure 29.5. Problem 29.17**

2. Label the current in the loops I

_{1}, I

_{2}, I

_{3}, ....., 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.

**Figure 29.6. Current loops in problem 29.17.**

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_{2}can be obtained by substituting eq.(29.11) into eq.(29.8):

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_{1}can be obtained from eq.(29.7):

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**which is based on**

__branch method__**:**

__Kirchhoff's first rule__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):

**Figure 29.7. Branch method applied to problem 29.17.**

2. Label the currents in each branch I

_{1}, I

_{2}, I

_{3}, ....., 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

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The current I

_{1}can be obtained from eq.(29.14):

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_{4}:

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_{3}can be obtained from eq.(), using the solution for I

_{4}just derived (eq.(29.21)):

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_{2}can be obtained from eq.(29.17):

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_{5}can be obtained from eq.(29.19)

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**: since the loop method involves fewer unknown and fewer equations, it is usually quicker to use than the branch method.**

__Note__### Example: Problem 29.10

Consider the circuit shown in Figure 29.8. Given that [epsilon]_{1}= 6.0 V, [epsilon]

_{2}= 10.0 V, and R

_{1}= 2.0 [Omega], what must be the value of the resistance R

_{2}if the current through this resistance is to be 2.0 A ?

**Figure 29.8. Problem 29.10.**

_{1}is equal to R

_{1}

^{.}(I

_{1}- I

_{2}). The sum of all emfs and all potential drops across the resistors in loop 1 is

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_{2}is thus equal to

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_{1}= 6.0 V, [epsilon]

_{2}= 10.0 V and I

_{2}= 2.0 A. These values combined with eq.(29.28) can be used to determine R

_{2}:

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## 29.4. Energy in circuits

**To 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**

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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

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**.**

__Joule heating__### Example: Example 8

A 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 10

^{5}kW of power at 2.3 x 10

^{5}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 10

^{5}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, P

_{delivered}, is equal to 1.7 x 10

^{5}kW. The voltage delivered, [Delta]V

_{delivered}, is equal to 2.3 x 10

^{5}V. The current through the cables can determined by applying eq.(29.32):

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b) If the voltage delivered to the city is 110 V, than current through the transmission cables must be equal to

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### Example: Problem 29.24.

A 12-V battery of internal resistance R_{i}= 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:

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_{1}= 12.0 V and it has an internal resistance R

_{i}= 0.20 [Omega]. If the recharger is to deliver a current I = 6.0 A then the emf of the recharger must be equal to

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**Figure 29.9. Problem 29.24.**

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### Example: Problem 29.32

Suppose 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 R

_{i}. The voltage drop across the internal resistance is equal to I R

_{i}. The external voltage of the battery if thus equal to ([epsilon]

_{b}- I R

_{i}). The power delivered by the battery to the external circuit is therefore equal to

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## 29.5. The RC circuit

**The 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

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**Figure 29.10. Simple RC circuit.**

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_{0}is the charge on the capacitor at time t = 0. After evaluating both integrals in eq.(29.52) we obtain

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_{0}= 0 C). Equation (29.55) can then be rewritten as

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_{f}on the capacitor is equal to

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