Sum1 Lec1 1 Resistive Cricuits & Equivalent Circuits

ESP1104 Summary Notes (1)

© Lim Fang Jeng

1

Lecture 1:

Resistive Networks and Circuits

Resistant,

A R_ ^

ρ= resistivity; ℓ= length of resistor; A=cross-section of resistor; Kirchoff’s Current Law

Sum of current of a node is zero. I1=I2+I3

Kirchoff’s Voltage Law

Sum of voltage of a closed loop is zero V1+V2-V3+V4=0

Potential Divider Current Divider

V₀ = ^RL + _�^�

I_L = ^RS

�^{+ IS}

In practical, when we use voltmeter and ammeter to measure the voltage and ampere. The result often has some sort of error due to the internal resistances.

For voltmeter, the internal resistance should be made large enough (10MΩ) so that it can potentially divide more voltage to the voltmeter and hence causing less error.

For ammeter, the internal resistance should be made as small as possible (1Ω) so that its resistor will not divide most of the voltage which will then cause larger error. By doing so, the error of reading will be minimized.

Node- Voltage Analysis

Equations of current can be formed by analyzing the node voltages

v₁= v_S v_{2− v1}

2

+^{v2− v3} R₃ ⁺

�2 4

= 0 v₃

R₅⁺

�3− �1 1

+^{�3− �2}

3

= 0

ESP1104 Summary Notes (1)

© Lim Fang Jeng

2

Thevenin & Norton equivalent circuits

Every circuits can be represented by one voltage source or current source and a resistor. Finding Thevenin’s Resistance

(1) Remove load (2) Zero all sources

(3) Compute the effective resistance of the circuit

Thevenin Voltage – Equivalent open circuit voltage which present at the load terminals Computing Thevenin’s Voltage

(1) Remove load resistance

(2) Define an open circuit voltage vOC

(3) Solve for vOC

(4) _{Set Vt} = v_OC

Norton’s Circuit – Replace the voltage source to a current source

Norton’s current – short-circuit current which present in the load terminals Finding Norton’s current

(1) Remove load, short the terminals joining the load (2) Define a short-circuit current iSC

(3) _{Set In} = i_SC

A Thevenin equivalent circuit can be converted to the Norton’s equivalent circuit

Maximum Power Transfer

Let the power dissipated by the load be P_L,

P_L = i_L²R_L

From the Thevenin’s equivalent circuit, we know that the load current, i_L is: i_L = ^Vt

R_T + R_L Substituting into the power dissipation equation,

P_L = ^Vt R_T + R_L

2

R_L

= ^Vt

2_R L

+ _� ² dP_L

� _� ⁼

⁺ _� �_�²− 2�_�² _�

+ _� ^{3 = 0}

R_L = R_T

Maximum power transfer will occur when the load resistance R_L is same as the Thevenin Resistance, R_T

ESP1104 Summary Notes (1)

© Lim Fang Jeng

3

Wheatstone Bridge

i₁ = i_g + i₃, i₂ + i_g = i₄ v_a = i₃R₃, v_b = i₄R_x

In null condition, i_g = 0, v_ab = 0, Hence,

i₁ = i₃, i₂ = i₄

v_a = v_b = i₃R₃ = i₄R_x i₁R₁ = i₂R₂ We obtain

R₃

1

= ^�

 2

R_x = R₂^R3 R₁ Deflection Bridge Circuit

We now fix R₁ = R₂, and R_x = R₃+_{∆R where R3} is to be varied. V_ba = V_{b − Va} = ^Rx

R₂+ R_x^{� −}

3 1⁺ 3^�