E84 Home Work 1

1. (a) If two light bulbs both labeled as 110V and 40W in series are connected to a socket outlet of 190V, what is the power consumption of each of the bulbs?

   Solution:
   

   $$R=V^2/W=110^2/40,\;\;\;\;\; V'=190/2=95$$
   
   
   
   
   $$W'=V'^2/R=W (V'/V)^2=40 (95/110)^2=40\times 0.746=29.83$$
   
   

   (b) Replace one of the two bulbs by another bulb labeled as 110V 15W, and find the power consumption of each of the bulbs. What will happen to each of the two bulbs? (Note that when the power consumption by a bulb is larger than the specified wattage, it will be burned out!)

   Solution:
   

   $$V=110,\;\;V'=190,\;\;R1=V^2/W1=110^2/40,\;\;R2=V^2/W2=110^2/15$$
   
   
   
   
   $$I=V'/R=V'/(R1+R2),\;\; W1'=I^2 R1=8.89,\;\;W2'=I^2 R2=23.67>15\;\;\; \mbox{burned out!}$$
   
   

2. Measurement of a physical process by instruments may be tricky as the instruments will inevitably affect the process being measured. The figure below shows two possible configurations for the measurement of the voltage across and the current through the load.

   

   1. What are required of the ammeter and the voltmeter to minimize their influences on the measurements?

      Solution: The ammeter should have minimum (ideally 0) impedance while the voltmeter should have maximum (ideally infinity) impedance.

   2. How would the ammeter and the voltmeter affect the measurement of the current and the voltage in either of the configurations (a and b)?

   Solution: In (a) the voltmeter will by-pass some current so that the actual current through the load is smaller than the reading of the ammeter.

   In (b) the ammeter will cause some voltage drop and the actual voltage across the load is lower than the reading of the voltmeter.

3. Use Kirchhoff's voltage and current laws to find voltage $V_{BD}$ and resistance $R_2$ in the circuit shown below:

   

   (Note: The direction of a current and the polarity of a voltage source can be assumed arbitrarily. To determined the actual direction and polarity, the sign of the values also should be considered. For example, a current labeled in left-to-right direction with a negative value is actually flowing right-to-left.)

   Solution:

   • apply KCL to node B: $2+(-6)+I2=0$, $I2=-4$
   • apply KCL to node C: $-4+1-I3=0$, $I3=-3$
   • apply KCL to node D: $5-3+I4=0$, $I4=-2$
   • apply KVL to loop ABCD: $10+2\times 1+(-4)\times R2+6+(-3)\times 1-(-2)\itimes 5=0$, $R2=6.25$
   • apply KVL to loop BDA: $V_{BD}+2\times 5+10+2\times 1=0$, $V_{BD}=-22V$, or
   • apply KVL to loop BDC: $V_{BD}-(-3)\times 1-6-(-4\times 6.25)=0$, $V_{BD}=-22V$

4. Find the equivalent resistance between the two terminals before and after the switch is closed. (Note, the two diagonal branches are NOT connected to each other in the middle.)

   

   Solution: before S is closed, $R=(3+2)//(6+2)=5//8=40/13 \Omega$

   after S is closed, $R=2//2+3//6=1+2=3 \Omega$

5. Find the equivalent resistance $R_{eq}$ between the two terminals in the figure, where $R_0=3\Omega$, $R_1=4\Omega$, $R_2=2\Omega$, $R_3=2\Omega$, $R_4=1\Omega$. What is $R_{eq}$ if $R_0=5\Omega$?

   (Hint: apply a test voltage $V_{test}$ across the terminals and the equivalent resistance can be found to be $R_{eq}=V_{test}/I_{test}$. The circuit can be solved by applying KCL to $V_1$ and $V_2$.)

   

   Solution:

   
   

   $$\frac{v_1-v_{test}}{4}+\frac{v_1-0}{2}+\frac{v_1-v_2}{3}=0$$
   
   
   
   
   $$\frac{v_2-v_{test}}{2}+\frac{v_2-0}{1}+\frac{v_2-v_1}{3}=0$$
   
   
   solving these we get $v_1=v_2=v_{test}/3$. Then apply KCL to the positive terminal to get
   
   $$i_{test}+\frac{v_1-v_{test}}{4}+\frac{v_2-v_{test}}{2}=0$$
   
   
   solving to get
   
   $$R_{eq}=\frac{v_{test}}{i_{test}}=2 \Omega$$
   
   
   Note that in this case $v_1=v_2$, independent of the value of $R_0$.

   Alternatively, based on delta-Y conversion (to be considered later), the triangle (delta) formed by $R_0$, $R1$, and $R_2$ can be coonverted to a Y configuration with $R'_0=8/9$ (top), $R'_1=4/3$ (left), and $R'_2=2/3$. Then we have $R'_1+R_3=10/3$ and $R'_2+R_4=5/3$. Their parallel combination is $10/3 \vert\vert 5/3=10/9$, in series with $R'_0=8/9$, i.e., the total resistance is $R_{eq}=10/9+8/9=2$. However, this approach does not reveal the fact that $R_0$ can take any value without changing $R_{eq}$.

6. Design a multimeter that can measure both DC and AC voltage, DC current, and resistance with different scales. Specifically, you are given an analog meter $A$ with a needle display, which reaches full scale when a DC current of $I=100\;\mu A=10^{-4}\;A$ goes through it. The internal resistance of the meter is 10 Ohms. In addition, you need some multi-position rotary switches to select different scales for each of the three types of measurements, and resistors with any values needed in your design.
   • DC Voltage measurement: DC voltages in these ranges can be measured 0-2.5, 0-10, 0-50, and 0-250 (all in volts). Use a 4-position rotary switch to select one of the four ranges as shown in the figure below. For example, when the range of 0-10 is selected, the needle display will reach full scale when the voltage being measured is 10 V. The circuit is shown below. Determine all resistances labeled.

     

     Solution: $R_1=25\;K\Omega$, $R_2=75\;K\Omega$, $R_3=400\;K\Omega$

     and $R_4=2\;M\Omega$.

   • AC Voltage measurement: To measure an AC voltage (in terms of its RMS value), it first needs to be converted into a DC voltage. This can be achieved by a diode which only allows the current to pass in one direction (along the arrow) but not the other. This process is called rectification. The diode will also cause a voltage drop of 0.7 volt along the direction. The actual reading of the meter reflects the average value of the rectified current. Find the resistance $R$ so that when the incoming AC voltage is $V=10$ volt (RMS), the meter shows a full scale display.

     

     Solution: The peak value of voltage is $10\times \sqrt{2}=14.14$, which is reduced (due to voltage drop of the diode) to $14.14-0.7=13.44$. The average value of the rectified voltage is $13.44/\pi=4.28$. For the meter to have a full scale display, the current need to be
     

     $$10^{-4}=\frac{4.28}{R},\;\;\;\;\;\;\mbox{i.e.}\;\;\;\;\;\;R=42.8\,k\Omega$$
     
     
     The internal resistance of $10\,\Omega$ can be neglected.

   • DC current measurement: measure currents in these ranges (all in mA): 0-0.5, 0-2.5, 0-10, 0-50. Use a 4-position rotary switch to select one of the four ranges as shown in the figure below. For example, when the range of 0-10 is selected, the needle display will reach full scale when a 10 mA current is measured. Determine all resistances labeled. Use $R_0=1\;K\Omega$.

     

     Solution: Voltage across input is $V=0.1\;mA\times 1\;K\Omega=100\;mV$. Therefore
     

     $$\begin{array}{l} R_1=100/(0.5-0.1)=250\;\Omega \\ R_2=100/... ...0.1)=10.1\;\Omega \\ R_1=100/(50-0.1)=2\;\Omega \end{array}$$
     
     

   • Resistance measurement: The circuit for resistance measurement is provided as shown below, where $V_1=1.5V$. Determine the values for the resistors labeled as $R_0$, $R_1$, $R_{10}$, $R_{100}$ and $R_{1000}$ and $V_2$ so that the needle display of the meter is full scale ($I=100\;\mu A$) when the resistor $R=0$ being measured (between the two leads labeled + and -) is zero, or half scale ($I=50\;\mu A$) when the value of $R$ and the position of he two synchronized rotary switches are given in each of the four case shown in the table:

     positions	$\times 1$	$\times 10$	$\times 100$	$\times 1K$
     $R$ values	20$\Omega$	200$\Omega$	2000$\Omega$	20 $K\Omega$

     

     Solution:

     • First determine $R_0$: when $R=0$, we get $R_0=1.5V/0.1\;mA=15\;K\Omega$.
     • When $R=20\;\Omega$, the current through meter $A$ should be:
       

       $$I = \frac{V_1}{R+R_1\vert\vert R_0}\frac{R_1}{R_1+R_0} =\frac{V_1}{R+\frac{R_1R_0}{R_1+R_0}}\frac{R_1}{R_1+R_0} =V_1\frac{R_1}{RR_1+RR_0+R_1R_0}$$

       $$= 5\times 10^{-4}\;A$$

       
       
       Given $V_1=1.5V$, $R_0=15K$ and $R=20\Omega$, we can solve this equation to get
       
       $$R_1=\frac{R_0RI}{V_1-I(R+R_0)}=20\Omega$$
       
       

     • When $R=200\;\Omega$, solving the above equation we get $R_{10}=202.7\Omega$.
     • When $R=2\;K\Omega$, solving the above equation we get $R_{10}=2307.7\Omega$.
     • When $R=20\;K\Omega$, we need to determine $V_2$ and $R_{1k}$ so that $I=5\items 10^{-5} A$, and also when $R=0$, $I=10^{-4}A$. $R_{1k}$ and $V_1+V_2$ can be found by solving these equations:
       
       $$\left\{ \begin{array}{l} (V_1+V_2)/(R_{1k}+R_0)=10^{-4}A \\ (V_1+V_2)/(R+R_{1k}+R_0)=5\times 10^{-5}A \end{array}\right.$$
       
       
       Solving we get $V_1+V_2=2V$, $R_{1k}+R_0=20\;K\Omega$, i.e., $R_{1k}=5K\Omega$, $V_2=0.5V$