基尔霍夫定律英文版ppt课件

Example 3.4. The potentiometer The Figure shows the rudiments of a potentiometer, which is a device for measuring an unknown emf x. The currents and emfs are marked as shown. Applying the loop theorem to loop abed yields Sr b a c d e R0 RS G 0 I I0 I0 I0 - I Galvanometer Adjust R0 to make I=0 I0RS=S Applications of Kirchhoffs Law Standard emf Adjust Rx to make I=0 I0RS=S I0Rx=x Applications of Kirchhoffs Law xr b a c d e R0 Rx G 0 R1 R2 R4 R3 a b G c d I0 I1 I3 Ig I1-Ig I3+Ig The Wheatstone bridge. In figure on the right ,R2 is to be adjusted in value until points a and b are brought to exactly the same potential. (One tests for this these points are at the same potential, the meter will not deflectbalanced.) Show that when this adjustment is made, the following relation holds: R2R3=R1R4 Solution: According to Kirchhoffs Law, We get Wheatstone bridge I0=I1+I3 I3R3+(I3+Ig)R4-=0 (I1-Ig)R2-(I3+Ig)R4-IgRg=0 I1R1+IgRg-I3R3=0 If (R2R3-R1R4)=0,then，Ig=0,No current passes through the Galvanometer, the bridge is in the balance. When a bridge is in the balance, The equivalent resistance between c and d can be calculated as follow (R1+R2)/ (R3+R4) or(R1/R3)+ (R2/R4) ？ The condition for balance of bridge: R2R3=R1R4 Cut circuit short circuit Wheatstone bridge 3.2.13 When a metal rod is heated, not only its resistance but also its length and its cross-sectional area change. The relation R = L/A suggests that all three factors should be taken into account in measuring at various temperatures. (a) If the temperature changes by 1.0C, what percent changes in R, L, and A occur for a copper conductor? (b) What conclusion do you draw? The coefficient of linear expansion is =1.710-5C. Solution: From =0(1+t) From l=l0(1+t) Solving Problems The logarithm of the equation above is Differential to the equation is So the change of with temperature is mainly part. Solving Problems 12 Very large Very large A B C II d1d2 S Find (a) the electric fields in the materials 1 and 2 ,E1and E2 (b) Potential differences VAB,VBC Solution: The materials are conducting From the definition of current density According to differential form of Ohms Law: j=E, we can get Solving Problems (c)Find the charges on the surfaces A,B,C. 12 Very large Very large A B C II d1d2 S Because is very large，the electric fields in the conductors, right and left, are zero Choose a Gaussian surface enclosing surface A Solving Problems xy (a) x (b) y I1 I1 I2 I2 I1-I2 Vxy=I1R+I2(2R) I I=I1+I2 ab Loop xabx:I1R+(I1-I2)R-2I2R=0I2/I1=2/3 Solving Problems R1 a b 1 2 3 R2 Find the current in each resistor and the potential difference between a and b. Put 1 = 6.0V, 2 = 5.0V, 3 = 4.0V, R1 = 100, and R2 =50 . Solution: Vab=2+3=9V I1 I2 I1R1-2=0 I1=2/R1=50mA I2R2+1-2-3=0 I2=(2+3 - 1) /R2=60mA Solving Problems 1,r1 2,r2 3,r3 R1 R2 R3 R4 R5 a b 1=12V, 2=9V, 3= 8V r1=r2=r3=1 R1=R3=R4=R5=2 R2=3 (a)ab is not connected ,find Vab; (b)ab is connected ,find Vab and the current flowing R2 Solution: 3-1+I(R1+R3+R4+R5+r1+r2)=0 Solving Problems 1,r1 2,r2 3,r3 R1 R2 R3 R4 R5 a b Vab=3- 2 +I(R3+R4+r3) =