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Solutions to problem 1-3 of electrical and computer engineering (ece) 002 lab homework #5 at the george washington university. The problems involve series and parallel circuit analysis, diode circuit analysis, and mesh current method. Students are required to calculate currents, voltages, and voltage drops using given resistor values and diode threshold voltage.
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Problem #1 Series Circuit Analysis (20 points) The Battery’s Voltage is: Vs = 6 V The Resistance of each resistor: R 1 = 100 Ω, R 2 = 200 Ω, R 3 = 300 Ω Calculate the current along this simple series circuit. Solution: I = (Don’t forget the units…amps, volts, ohms…or the prefix: milli, micro, nano, pico)
Problem #2 Parallel Circuit Analysis(20 points) The Resistance of each resistor: R 1 = 200 Ω, R 2 = 300 Ω, The Current across each resistor: I 1 = 60 mA , I 2 = 40 mA Calculate the voltage across each resistor VR1 and VR2. Calculate the battery’s voltage: Vs? Solution: VR1= VR2= VS = (Don’t forget the units…amps, volts, ohms…or the prefix: milli, micro, nano, pico)
DC R 1 = 1 KΩ Vth= 0. 7 V R 2 = 1 KΩ R 3 = 1 KΩ Using the above resistor values and a DC voltage of 3.7V, calculate the total current flowing through the circuit. The diode’s threshold voltage is 0.7V (Voltage required to turn the diode “on”). Solution: Itotal= (Don’t forget the units…amps, volts, ohms…or the prefix: milli, micro, nano, pico) Itotal
Problem #5 Mesh Current Analysis – Why we need Matlab in Circuits (30 points) Taking a look at this circuit, it is called unbalanced Wheatstone bridge. It is a famous and very practical circuit configuration used in many devices today: Figure 1 Battery: Vs = 24 V Resistors: R 1 = 150Ω, R 2 = 50 Ω, R 3 = 100 Ω, R 4 = 300 Ω and R 5 = 250 Ω This type of circuit is irreducible by normal series-parallel analysis. You can see that nothing is in parallel nor in series with anything else. Try it out if you don’t believe! One way to find the current in this circuit, is to use something called the “Mesh Current Method”, which we full explain below. STEP 1 The first step in the ‘Mesh Current’ method is to draw something called the “mesh currents” to account for all the possible paths current can ‘flow’ in the circuit. Look at figure 2, to see how we drew the ‘mesh currents’. The directions of these mesh currents, are arbitrary, in the end, if we were wrong in the direction: I1, I2, or I3 will have a negative sign telling us to reverse the arrow’s direction. Why are there 3 currents? Two mesh currents (I1 & I2) are not enough in this circuit, because neither I 1 nor I 2 goes through the battery. So, we must add a third mesh current, I 3. Here, we have chosen I 3 to loop from the bottom side of the battery, through R 4 , through R 1 , and back to the top side of the battery. This is not the only path we could have chosen for I 3 , but it seems the simplest.
The voltage drop across R1: VR1 = R 1 (I 1 + I 3 ) The sum of the voltage drops around this loop = VR1+VR2 + VR Therefore, from KΩVL: VR1+VR2 + VR3 = 0 => Substituting the expressions for VR1,VR2 and VR3 into the above equation: R 2 I 1 + R 3 (I 1 + I 2 ) + R 1 (I 1 + I 3 ) = 0 => Substituting the values of R 1 , R 2 and R 3 into the above equation: 50 I 1 + 100(I 1 +I 2 ) + 150(I 1 +I 3 ) = 0 => Simplifying the above equation: 300 I 1 + 100 I 2 +150 I 3 = 0 (1) ****NOTE 1 : In the equation above: VR3 = R 3 (I 1 + I 2 ), why did we ‘add’ I1 and I2? Why not take the difference? Because mesh current I1 and mesh current I2 flow in the ‘same’ direction through resistor R3, so it makes sense to add them! If they were in opposite directions, we would subtract them. How do we know which one to subtract? It depends on what loop we’re in! If we’re in I1’s loop, we’d subtract I2 (if I2 were flowing the other way) FOR THE BOTTOM LOOP ( where I2 is located in figure 2 ): The voltage drop across R 3 : VR3 = R 3 (I 1 + I 2 ) The voltage drop across R 4 : VR4 = R 4 (I 2 – I 3 ) The voltage drop across R 5 : VR5 = R 5 I 2 The sum of the voltage drops around this loop = VR3 + VR4 + VR Therefore, from KΩVL (and simplifying a bit): VR3 + VR4 + VR5= 0 => R 3 (I 1 + I 2 ) + R 4 (I 2 – I 3 ) + R 5 I 2 = 0 => 100 (I 1 + I 2 ) + 300 (I 2 – I 3 ) + 250 I 2 = 0 => 100 I 1 + 650 I 2 – 300 I 3 = 0 (2) FOR THE LEFT LOOP ( Containing the battery & mesh current I3 ): The voltage drop across the battery: Vs = 24 V The voltage drop across R1: VR1 = R 1 (I 1 + I 3 ) The voltage drop across R4: VR4 ’= -VR4 = - R 4 (I 2 – I 3 ) ****see note 2 below
The sum of the voltage drops around this loop = - Vs + VR1 + VR4’ Therefore, from KΩVL:
Your result: I =
. So that I 1 = ______________ I 2 = ______________ I 3 = ______________ (Do not forget write down the unit for the currents!) Bonus Question (10 points): Is the value of I 1 negative or positive? If I 1 is negative, what does this mean? Solution:
