了s=了is (7.2.2) neg g neg where f is the source force per unit charge.Outside the battery f=0,so we extend the path in Eq.(7.2.2)to the entire loop.In that case,the work done per unit charge by the non-electrostatic force,F,around a closed path is commonly referred to as the electromotive force,or emf (symbol s). i-j语6-js (7.2.3) path This is a poor choice of name because it is not a force but work done per unit charge.The SI unit for emf is the volt (V). Inside our ideal battery without any internal resistance,the sum of the electrostatic force and the source force on the charge is zero, qE+qf=0 (7.2.4) Therefore the electrostatic field is equal in magnitude to the source force per unit charge but opposite in direction, E=-f. (7.2.5) The electric potential difference between the terminals is defined in terms of the electrostatic field r(+)-(-)=-∫Es=∫is=e (7.2.6) The potential difference Al between the positive and the negative terminals of the battery is called the terminal voltage,and in this case is equal to the emf. Electromotive force is not restricted to chemical forces.In Figure 7.2.2,the inner working of a Van de Graaff generator are displayed.An electric motor drives a non- conducting belt that transports charge carriers in a direction opposite the electric field. The positive charge carriers are moved from lower to higher potential,and negative charge carriers are moved from higher to lower potential.Strong local fields at the brushes of the terminals both add and remove charge carriers from the belt.An electric motor provides the energy to move the belt and hence is the source of the electromotive force. 7-6
7-6 Fs q ⋅ d s neg pos ∫ = fs ⋅ d s neg pos ∫ , (7.2.2) where fs is the source force per unit charge. Outside the battery fs = 0 , so we extend the path in Eq. (7.2.2) to the entire loop. In that case, the work done per unit charge by the non-electrostatic force, Fs , around a closed path is commonly referred to as the electromotive force, or emf (symbol ε ). ε ≡ fs ⋅ d s closed path ∫ = Fs q ⋅ d s − + ∫ = fs ⋅ d s − + ∫ . (7.2.3) This is a poor choice of name because it is not a force but work done per unit charge. The SI unit for emf is the volt (V). Inside our ideal battery without any internal resistance, the sum of the electrostatic force and the source force on the charge is zero, q E + q fs = 0 . (7.2.4) Therefore the electrostatic field is equal in magnitude to the source force per unit charge but opposite in direction, E = − fs . (7.2.5) The electric potential difference between the terminals is defined in terms of the electrostatic field V (+) −V (−) = − E⋅ d s − + ∫ = fs ⋅ d s − + ∫ = ε . (7.2.6) The potential difference ΔV between the positive and the negative terminals of the battery is called the terminal voltage, and in this case is equal to the emf . Electromotive force is not restricted to chemical forces. In Figure 7.2.2, the inner working of a Van de Graaff generator are displayed. An electric motor drives a nonconducting belt that transports charge carriers in a direction opposite the electric field. The positive charge carriers are moved from lower to higher potential, and negative charge carriers are moved from higher to lower potential. Strong local fields at the brushes of the terminals both add and remove charge carriers from the belt. An electric motor provides the energy to move the belt and hence is the source of the electromotive force
V=V V=0 Figure 7.2.2 Van de Graaff generator Solar cells and thermocouples are also examples of emf source.They can also be thought of as a"charge pump"that moves charges from lower potential to higher potential. Consider a simple circuit consisting of a battery as the emf source and a resistor of resistance R,as shown in Figure 7.2.3 D R 0=-∫Es=0=-jEs-jEs leg 2 leg 1 Figure 7.2.3(a)Electric potential Figure 7.2.3(b)Circuit diagram. difference for leg 1 and leg 2 sum to zero. The circuit diagram in Figure 7.2.3(b)corresponds to the circuit in Figure 7.2.3(a).The electric potential difference around the loop is zero because the electrostatic field is conservative,Eq.(7.2.1).We can divide the loop into two legs;leg 1 goes from the positive terminal to the negative terminal through the external circuit,and leg 2 goes from the negative terminal to the positive terminal through the battery, 0=-∫Es=0=-jEs-jEs (7.2.7) oop The integral via leg 1 in the external circuit is just the potential difference across the resistor,which is given by Ohm's Law,where we have assumed that the wires have negligible resistance, 7-7
7-7 Figure 7.2.2 Van de Graaff generator Solar cells and thermocouples are also examples of emf source. They can also be thought of as a “charge pump” that moves charges from lower potential to higher potential. Consider a simple circuit consisting of a battery as the emf source and a resistor of resistance R, as shown in Figure 7.2.3. Figure 7.2.3(a) Electric potential difference for leg 1 and leg 2 sum to zero. Figure 7.2.3(b) Circuit diagram. The circuit diagram in Figure 7.2.3(b) corresponds to the circuit in Figure 7.2.3(a). The electric potential difference around the loop is zero because the electrostatic field is conservative, Eq. (7.2.1). We can divide the loop into two legs; leg 1 goes from the positive terminal to the negative terminal through the external circuit, and leg 2 goes from the negative terminal to the positive terminal through the battery, 0 = − E⋅ d s = 0 loop ∫ = − E⋅ d s 1 + − ∫ − E⋅ d s2 − + ∫ , (7.2.7) The integral via leg 1 in the external circuit is just the potential difference across the resistor, which is given by Ohm’s Law, where we have assumed that the wires have negligible resistance
△r=-∫E.d=-lIR (7.2.8) The integral via leg 2 through the battery is the emf (Eq.(7.2.6), An--jE.di=e. (7.2.9) Eq.(7.2.7)becomes 0=△V+△V,=-IR+e. (7.2.10) Therefore the current in the loop is given by I=e (7.2.11) R However,a real battery always carries an internal resistance r(Figure 7.2.4a),and the potential difference across the battery terminals becomes △V=e-Ir. (7.2.12) d Ir IR b Figure 7.2.4 (a)Circuit with an emf source having an internal resistance r and a resistor of resistance R.(b)Change in electric potential around the circuit. Because there is no net change in potential difference around a closed loop,we have 8-Ir-IR=0. (7.2.13) Therefore the current through the circuit is 7-8
7-8 ΔV1 = − E⋅ d s 1 + − ∫ = −IR . (7.2.8) The integral via leg 2 through the battery is the emf (Eq. (7.2.6), ΔV2 = − E⋅ d s − + ∫ = ε . (7.2.9) Eq. (7.2.7) becomes 0 = ΔV1 + ΔV2 = −IR + ε . (7.2.10) Therefore the current in the loop is given by I R ε = . (7.2.11) However, a real battery always carries an internal resistance r (Figure 7.2.4a), and the potential difference across the battery terminals becomes ΔV = ε − Ir . (7.2.12) Figure 7.2.4 (a) Circuit with an emf source having an internal resistance r and a resistor of resistance R. (b) Change in electric potential around the circuit. Because there is no net change in potential difference around a closed loop, we have ε − Ir − IR = 0 . (7.2.13) Therefore the current through the circuit is
I= (7.2.14) R+r Figure 7.2.4(b)depicts the change in electric potential as we traverse the circuit clockwise.From the figure,we see that the highest potential is immediately after the battery.The potential drops as each resistor is crossed.Note that the potential is essentially constant along the wires.This is because the wires have a negligibly small resistance compared to the resistors. 7.3 Electrical Energy and Power Consider a circuit consisting of an ideal battery (zero internal resistance)and a resistor with resistance R(Figure 7.3.1).The potential difference between two points a and b be g=V-V>0.If a charge Ag is moved through the battery,its electric potential energy is increased by AU=Age.On the other hand,as the charge moves across the resistor, the potential energy is decreased due to collisions with atoms in the resistor.If we neglect the internal resistance of the battery and the connecting wires,upon returning to a,the change in potential energy of Ag is zero. R Figure 7.3.1 A circuit consisting of an ideal battery with emf g and a resistor of resistance R. The rate of energy loss through the resistor is given by P= △U △ (7.3.1) △1△1 This is equal to the power supplied by the battery.Using &=IR in Eq.(7.3.1),one may rewrite the rate of energy loss through the resistor as P=IR (7.3.2) Using I=8/R in Eq.(7.3.1),the power delivered by the battery is P=82/R (7.3.3) 7-9
7-9 I R r ε = + . (7.2.14) Figure 7.2.4(b) depicts the change in electric potential as we traverse the circuit clockwise. From the figure, we see that the highest potential is immediately after the battery. The potential drops as each resistor is crossed. Note that the potential is essentially constant along the wires. This is because the wires have a negligibly small resistance compared to the resistors. 7.3 Electrical Energy and Power Consider a circuit consisting of an ideal battery (zero internal resistance) and a resistor with resistance R (Figure 7.3.1). The potential difference between two points a and b be ε =Vb −Va > 0 . If a charge Δq is moved through the battery, its electric potential energy is increased by ΔU = Δqε . On the other hand, as the charge moves across the resistor, the potential energy is decreased due to collisions with atoms in the resistor. If we neglect the internal resistance of the battery and the connecting wires, upon returning to a, the change in potential energy of Δq is zero. Figure 7.3.1 A circuit consisting of an ideal battery with emf ε and a resistor of resistance R. The rate of energy loss through the resistor is given by P = ΔU Δt = Δq Δt ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ε = Iε . (7.3.1) This is equal to the power supplied by the battery. Using ε = IR in Eq. (7.3.1), one may rewrite the rate of energy loss through the resistor as P = I 2 R . (7.3.2) Using I = ε / R in Eq. (7.3.1), the power delivered by the battery is P = ε2 / R. (7.3.3)
a Figure 7.3.2 A circuit consisting of a battery with emf g and a resistor of resistance R. For a battery with emf g and internal resistance r(Figure 7.3.2),the power or the rate at which chemical energy is delivered to the circuit is P=I8=1(IR+Ir)=IR+Ir. (7.3.4) The power of the source emf is equal to the sum of the power dissipated in both the internal and load resistance as required by energy conservation. 7.4 Resistors in Series and in Parallel The two resistors with resistance R and R,in Figure 7.4.1 are connected in series to a source of emf g.By current conservation,the same current,I,is in each resistor. R W Rea Figure 7.4.1 (a)Resistors in series.(b)Equivalent circuit. The total voltage drop from a to c across both elements is the sum of the voltage drops across the individual resistors: e=△V=IR+IR=I(R+R) (7.4.1) 7-10
7-10 Figure 7.3.2 A circuit consisting of a battery with emf ε and a resistor of resistance R. For a battery with emf ε and internal resistance r (Figure 7.3.2), the power or the rate at which chemical energy is delivered to the circuit is 2 2 P = Iε = I(IR + Ir) = I R + I r . (7.3.4) The power of the source emf is equal to the sum of the power dissipated in both the internal and load resistance as required by energy conservation. 7.4 Resistors in Series and in Parallel The two resistors with resistance R1 and R2 in Figure 7.4.1 are connected in series to a source of emf ε . By current conservation, the same current, I , is in each resistor. Figure 7.4.1 (a) Resistors in series. (b) Equivalent circuit. The total voltage drop from a to c across both elements is the sum of the voltage drops across the individual resistors: ε = ΔV = I R1 + I R2 = I(R1 + R2 ) . (7.4.1)