4-5基尔霍夫定律英文版

If (R2R3-R1R4)=0,then，Ig=0,No current passes through the Galvanometer, the bridge is in the balance. The condition for balance of bridge: R2R3=R1R4 When a bridge is in the balance, The equivalent resistance between c and d can be calculated as follow
Standard emf
ε x + I (r + RG ) − (I0 − I )Rx = 0 b I0 Rx − ε x
I= R + r + RG
a Adjust Rx to make I=0
R0
εx
r
G
c e d
Rx
ε0
I0RS=εS I0Rx=εx
Rx ε x = εs Rs
Wheatstone bridge
R2
a
ε2
R1
ε3
b
I2
I1
I1=ε2/R1=50mA I2=(ε2+ε3 - ε1) /R2=60mA
Solving Problems
ε1=12V, ε2=9V, ε3= 8V r1=r2=r3=1Ω R2=3Ω
ε1,r1 a b ε2,r2 ε3,r3 R2 R5 R4
R1 R3
R1=R3=R4=R5=2Ω (a)ab is not connected ,find Vab;
ε1,r1 a b ε2,r2 I2 ε3,r3 R2 I R5 R4
II
(b) When ab is connected, from Kirchhoff’s Law
ε2-ε3+I3(r3+R3+R4)+I2(r2+R2)=0
l R=ρ A
The logarithm of the equation above is
ln R = ln ρ + ln l − ln A
Differential to the equation is
∆R ∆ρ ∆l ∆A = + − = 0.39% + 0.0017% − 0.0034% = 0.3883% R ρ l A
ρ − ρ0 = α∆t = 3.9 ×10 −3 ×1 = 0.39% ρ0
From l=l0(1+γ∆t)
l − l0 = γ∆t = 1.7 ×10 −5 ×1 = 0.0017% l0
Solving Problems
A − A0 = 2γ∆t = 2 ×1.7 × 10 −5 × 1 = 0.0034% A0
Solving Problems
ε1
Find the current in each resistor and the potential difference between a and b. Put ε1 = 6.0V, ε2 = 5.0V, ε3 = 4.0V, R1 = 100Ω, and R2 =50 Ω. Solution: Vab=ε2+ε3=9V I1R1-ε2=0 I2R2+ε1-ε2-ε3=0
2 3 1 4
I1-Ig
Solution: According to Kirchhoff’s Law, We get I0=I1+I3 I3R3+(I3+Ig)R4-ε=0 I1R1+IgRg-I3R3=0 (I1-Ig)R2-(I3+Ig)R4-IgRg=0
Wheatstone bridge
Ig = ( R2 R3 − R1R4 ) R3ε R1 ( R3 + R4 )(Rg R4 + R2 R3 + R3 R4 + R3 Rg ) + ( R2 R3 − R1R4 )(R3 R4 + Rg R3 + Rg R4 )
a
The Wheatstone bridge. In figure on the R1 R2 I1 right ,R2 is to be adjusted in value until Ig points a and b are brought to exactly d c G the same potential. (One tests for this I3 R4 R3 I0 these points are at the same potential, I3+Ig b the meter will not deflect——balanced.) Show that when this adjustment is made, the following relation holds: R R =R R ε
( R1 + R2 )( R3 + R4 ) Rcd = ( R1 + R2 + R3 + R4 )
(R1+R2)// (R3+R4) Cut circuit
？
or
R1 R3 R2 R4 Rcd = + R1 + R3 R2 + R4
(R1//R3)+ (R2//R4) short circuit
Solving Problems
Applications of Kirchhoff’s Law
Example 3.4. The potentiometer The Figure shows the
rudiments of a potentiometer, which is a device for measuring an unknown emf εx. The currents and emfs are marked as shown. Applying the loop theorem to loop abed yields
εS + I (r + RG ) − (I0 − I )RS = 0
I0 RS − ε S I= RS + r + RG
Adjust R0 to make I=0 b I a I0
R0
εS
r
G
Galvanometer c e d I0
RS I0 - I
I0RS=εS
ε0
Applications of Kirchhoff’s Law
Solving Problems
l 3.5 × 10 −5 × 2 ×10 −2 R=ρ = = 8.9 × 10 −3 Ω A 3.14 × (0.5 × 10 − 2 ) 2
P=i2R
i= P 1 = = 1.06 A −3 R 8.9 × 10
i 1.06 j= = = 1.35 × 10 4 A / m 2 S 3.14 × (0.5 × 10 − 2 ) 2
A B C S I σ1 σ2 d1 d2
σ Very large
I
E=
j
σ
I
1
Choose a Gaussian surface enclosing surface A
(S )
∫∫ E ⋅ d S = E S = σ
1= qAຫໍສະໝຸດ / ε 0ε0I qA = σ1
ε0I qC = − σ2
ε0I ε0I qB = − σ 2 σ1
V=iR=1.06×8.9×10-3=9.434×10-3V
Solving Problems
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
Solving Problems y
R R Ra = + = R 2 2
I2
Ra = 2 R // 2 R = R x
y
I1
a
I1 I1-I2
b
Vxy=I1R+I2(2R) I=I1+I2 (a) x
I I2 (b)
I2 2 R + 2R R + 2R × Vxy I1 R + 2 I 2 R I1 3 =7R = Rxy = = = 2 I2 5 I I1 + I 2 1+ 1+ 3 I1 Loop xabx: I1R+(I1-I2)R-2I2R=0 I2/I1=2/3
Id2 VBC = E2d2 = σ 2S
Solving Problems
(c)Find the charges on the surfaces A,B,C. Because σ is very large，the electric fields in the conductors, right and left, are zero
j = I S
A B C I σ1 σ2 d1 d2
σ Very large
S
I
According to differential form of Ohm’s Law: j=σE, we can get j I I E1 = E2 = = σ 1 σ 1S σ 2S
Id1 VAB = E1d1 = σ1S
(b)ab is connected ,find Vab and the current flowing R2 Solution:
ε3-ε1+I(R1+R3+R4+R5+r1+r2)=0