Basic Electricity and Electronics Lecture PDF

 Lecture Outlines  Practical Perspective such as Resistive Touch Screen  Resistors in Series  Resistors in Parallel  Voltage Divider and Current Divider  Voltage Division and Current Division Resistive Touch Screen  Mobile phones and tablet computers use res...

 Lecture Outlines  Practical Perspective such as Resistive Touch Screen  Resistors in Series  Resistors in Parallel  Voltage Divider and Current Divider  Voltage Division and Current Division Resistive Touch Screen  Mobile phones and tablet computers use resistive touch screen. Two screens are used and seperated by insulating layer.  Touch screen is modeled by a grid of resistors in x-direction and Y-direction as shown in the figure on the right.  Electronic circuit applies a voltage across the grid in X-direction between points ‘a’ and ‘b’, then removes that Voltage and applies a voltage across the grid in Y-direction between points ‘c’ and ‘d’ and continuous to repeat this process.  When the screen is touched, the two screens are pressed together and create a voltage in a point of x-grid and y-grid to locate the point where the screen was touched Resistors in Series To find 𝑖𝑠 , we apply Kirchhoff’s Voltage Law 𝑣𝑠 ∴ 𝑖𝑠 = 𝑅𝑒𝑞 Combining Resistors in Series Resistors in Parallel To find 𝑖𝑠 , we apply Kirchhoff’s Current Law The resistors are connected in parallel, so the voltage across the resistors must be the Same. 𝑣𝑠 𝑣𝑠 𝑣𝑠 𝑣𝑠 ∵ 𝑖1 = & 𝑖2 = & 𝑖3 = & 𝑖4 = 𝑅1 𝑅2 𝑅3 𝑅4 ∴  Equivalent Resistance and Equivalent Conductance Sometimes, using conductance when dealing with resistors connected in parallel as follows: Example-1: Find 𝒊𝒔 , 𝒊𝟏 and 𝒊𝟐 in the circuit shown in Fig.1 Solution 3Ω and 6Ω are in series , so 3Ω+ 6Ω = 9Ω as shown in Fig.2 Fig.1 Also, 9Ω and 18Ω are in parallel, so 𝟏𝟖×𝟗 = 6Ω as shown in Fig.3 𝟏𝟖+𝟗 𝒗𝒔 𝟏𝟐𝟎 ∴ 𝒊𝒔 = = = 𝟏𝟐𝑨 Fig.2 𝑹𝒆𝒒 𝟏𝟎 𝒗𝟏 = 𝒊𝒔 × 𝑹𝟔Ω =12 × 6 = 72V 𝒗𝟏 𝟕𝟐 𝒊𝟏 = = = 4A 𝑹𝟏𝟖Ω 𝟏𝟖 𝒗𝟏 𝟕𝟐 𝒊𝟐 = = = 4A Fig.3 𝑹𝟗Ω 𝟗  The Voltage-Divider Voltage-Divider circuit is used to develop more than voltage level from a single voltage supply as shown in Fig. 4 Using Kirchhoff’s Voltage Law and Ohm’s Low Using Ohm’s Law Fig. 4 Example-2: the resistors in the voltage-divider circuit shown in Fig. 5 have a tolerance of ±10%. Find the maximum and minimum value of 𝒗𝟎 Solution the maximum value of 𝒗𝟎 is obtained by Fig. 5 increasing the value of 𝑹𝟐 and decreasing the value of 𝑹𝟏 𝟏𝟎𝟎 𝑹𝟐_𝒉𝒊𝒈𝒉 = 100 + = 𝟏𝟏𝟎 kΩ 𝟏𝟎 𝟐𝟓 𝑹𝟏_𝒍𝒐𝒘 = 25 - = 22.5 kΩ 𝟏𝟎 𝒗𝒔 ×𝑹𝟐 𝟏𝟎𝟎×𝟏𝟏𝟎 𝑽𝟎_𝒎𝒂𝒙 = = = 83.02 V 𝑹𝟏 +𝑹𝟐 𝟏𝟏𝟎+𝟐𝟐.𝟓 the minimum value of 𝒗𝟎 is obtained by decreasing the value of 𝑹𝟐 and increasing the value of 𝑹𝟏 𝟏𝟎𝟎 𝑹𝟐_𝒍𝒐𝒘 = 100 - = 𝟗𝟎 kΩ 𝟏𝟎 𝟐𝟓 𝑹𝟏_𝒉𝒊𝒈𝒉 = 25 + = 27.5 kΩ 𝟏𝟎 𝒗𝒔 ×𝑹𝟐 𝟏𝟎𝟎×𝟗𝟎 𝑽𝟎_𝒎𝒊𝒏 = = = 76.6 V 𝑹𝟏 +𝑹𝟐 𝟗𝟎+𝟐𝟕.𝟓  The Current-Divider Current-Divider circuit is used to divide the current between the parallel resistors as shown in Fig. 6 The voltage across the resistors in parallel is Fig. 6 Example-3: Find the power dissipated in the 6 Ω resistor as shown in Fig. 7 Solution 6 Ω and 4 Ω resistors are connected parallel Fig. 7 with 1.6 Ω series, so the total equivalent resistance of these resistors is 4 Ω as shown in Fig. 8 𝟒×𝟔 𝑹𝒆𝒒 = 1.6 + =4Ω 𝟒+𝟔 𝟏𝟔 𝒊𝟎 = 10 = 8A Fig. 8 𝟏𝟔+𝟒 𝒊𝟎 is the current in the 1.6 Ω 𝟒 𝒊𝟔Ω = 8 = 3.2 A ∴ P= 𝒊𝟐 𝑹𝟔 = 𝟑. 𝟐𝟐 (6)= 61.44 W 𝟔+𝟒  Voltage Division Voltage Division is a general result from analyzing the voltage divider as shown in Fig. 9 Using Ohm’s Low, the voltage 𝒗j across the resistor 𝑹j Fig. 9 Voltage Division Equation  Current Division Current Division is a general result from analyzing the current divider as shown in Fig. 10 Fig. 10 Using Ohm’s Low, the voltage 𝒗 across the resistors in parallel Current Division Equation Example-4: Use the current division to find 𝒊𝟎 and use voltage division to find the voltage 𝒗𝟎 for the circuit in Fig. 11 Solution Fig. 11 the voltage drop across the branch containing the 40Ω, the 10Ω, and the 30Ω resistors in series. We can then use voltage division to determine the voltage drop 𝒗𝟎 across the 30 Ω resistor. The voltage drop across the Fig. 11 series-connected resistors is 𝒗. Also, we need to calculate the equivalent resistance of the series-connected resistors that is equal to 40 + 10 + 30 = 80 Ω 𝒗 𝑹𝟑𝟎Ω 𝟒𝟖 × 𝟑𝟎 ∴ 𝒗𝟎 = = = 𝟏𝟖𝑽 𝑹𝒆𝒒 𝟖𝟎  References 1-James W. Nilsson, Susan A. Riedel, ‘’ Electric Circuits’’-tenth edition 2-Charles k. Alexander, Matthew N.O. Sadiku, ‘’ Fundamentals of Electric Circuits’’ – fourth edition