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Calculate The Effective Resistance Add in library

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Question:

1(a) Resistors in series Calculate the effective resistance for
 
Where: R1 = 2 Ω
R2 = 4 Ω
R3 = 9 Ω.
 
1(b) Resistors in parallel Calculate the effective resistance for:
where R = 6 Ω.
 
1(c) Three resistors in parallel Calculate the effective resistance for
where R1 = 2 Ω,
R2 = 6 Ω and
R3 = 9 Ω.
 
1(d) 2 resistors in parallel with 2 resistors in series
Calculate the effective resistance for the following:
 
where R1 = 2 Ω,
R2 = 4 Ω,
R3 = 4 Ω and
R4 = 5 Ω.
 
2(a) Analyse 2 parallel resistors in series with 1 other resistor In the circuit diagram shown in the figure,
 
where R1 =1 Ω,
R2 =5 Ω and
R3 =4 Ω
calculate the current drawn from the battery.
 
2(b) Analyse 2 parallel resistors in series with 1 other resistor In the circuit diagram shown in the figure for Q2(a),
where R1 =1 Ω,
R2 =5 Ω and
R3 =4 Ω. calculate the current through R1.
 
2(c) Analyse 2 parallel resistors in series with 1 other resistor In the circuit diagram shown in the figure,
 
here R1 =1 Ω, R2 =5 Ω and R3 =4 Ω. calculate the potential difference across resistor R3.
 
2(d) Analyse 2 parallel resistors in series with 1 other resistor In the circuit diagram shown in the figure,
 
where R1 =1 Ω, R2 =5 Ω and R3 =4 Ω. calculate the power dissipated in R2.
 
3 Galvanometer used as an ammeter & voltmeter A galvanometer has a full scale deflection for a current of 2 mA. The coil has a resistance of 100 Ω. Modify the instrument to produce the following.
(a) An ammeter that can measure up to 5 A.
(b) A voltmeter capable of measuring 10 V
 
 

Answer:

1(a) When resistors are connected in series their equivalent effective resistance is the sum of all these resistances. So, we have
Reff = 2 Ω + 4 Ω + 9 Ω
       = 15 Ω
 
1(b) When resistors are connected in parallel their equivalent effective resistance is the sum of the reciprocals of all these resistances. So, we have
1/Reff = (1/6) + (1/6)
          = (1 + 1) / 6
          = 2/6
          = 1/3 Ω
 
1(c) Similar to the above question, when resistors are connected in parallel their equivalent effective resistance is the sum of the reciprocals of all these resistances. So, we have
1/Reff = (½) + (1/6) + (1/9)
          = {(3 + 1) / 6)} + (1/9)
          = (4/6) + (1/9) = (2/3) + (1/9)
          = (6 + 1)/9
1/Reff = (7/9)
   Reff = (9/7)
   Reff = 1.285 Ω
 
1(d) First we solve the parallel combination of resistors and then their equivalent effective resistance is the sum of all these resistances. Let Ra is the parallel equivalent of R1 & R2
1/Ra = (½) + (1/4)
        = (2+1)/4
        = ¾
   Ra = 4/3
        = 1.33 Ω
Now, Ra, R3 & R4 are in series, so their equivalent effective resistance is
Reff = Ra + R3 + R4
       = 1.33 + 4 + 5
Reff = 10.33 Ω
 
2(a) istance is the sum of both these resistances i.e. Let Ra is the parallel equivalent of R1 & R2
1/Ra = (1/1) + (1/5)
        = (5+1)/5 1/Ra
        = (6/5)
   Ra = 5/6 Ra
        = 0.833 Ω
Now, Ra & R3 are in series, so their equivalent effective resistance is
 Reff = Ra + R3
        = 0.833 + 4
Reff  = 4.833 Ω
To find the current drawn from the battery we use ohm’s law
V = I Reff
I = V/Reff
I = 12/4.833
I = 2.4829 Amp
 
 
2(b) First we solve the parallel combination of resistors R1 & R2 i.e. Let Ra is the parallel equivalent of R1 & R2
1/Ra = (1/1) + (1/5)
        = (5+1)/5 1/Ra
        = (6/5) Ra = 5/6 Ra
        = 0.833 Ω
Now we apply the voltage division rule to find voltage drop across Ra i.e.
Va = {Ra / (Ra + R3)} x Vbattery
Va = {0.833 / (0.833 + 4)} x 12
Va = {0.833 / 4.833} x 12
Va = 0.1723 x 12 Va = 2.068 Volts
Current across R1 will be
Va = IR1 x R1
IR1 = Va / R1
IR1 = 2.068 / 1
IR1 = 2.068 Amp
 
2(c) First we solve the parallel combination of resistors R1 & R2 i.e. Let Ra is the parallel equivalent of R1 & R2
1/Ra = (1/1) + (1/5)
        = (5+1)/5
1/Ra = (6/5) Ra
        = 5/6 Ra
        = 0.833 Ω
Now we apply the voltage division rule to find voltage drop across R3 i.e.
VR3 = {R3 / (Ra + R3)} x Vbattery
VR3 = {4 / (0.833 + 4)} x 12
VR3 = {4 / 4.833} x 12
VR3 = 0.8276 x 12
VR3 = 9.93 Volts
 
2(d) First we solve the parallel combination of resistors R1 & R2 i.e. Let Ra is the parallel equivalent of R1 & R2
1/Ra = (1/1) + (1/5)
        = (5+1)/5 1/Ra
        = (6/5) Ra
        = 5/6 Ra
        = 0.833 Ω
Now we apply the voltage division rule to find voltage drop across Ra i.e.
Va = {Ra / (Ra + R3)} x Vbattery
Va = {0.833 / (0.833 + 4)} x 12
Va = {0.833 / 4.833} x 12
Va = 0.1723 x 12 Va = 2.068
Volts Power dissipated across R2 will be
PR2 = Va 2 /R2
PR2 = (2.068)2 / 5
PR2 = 4.276 / 5
PR2 = 0.855 Watts
 
3(a)  For a given meter, the deflection is proportional to the coil current. The corresponding coil voltage is
VT = Rm Imf
     = 100 (0.002)
     = 200 mV
Rm is the movement resistance, and Imf is the current required for full scale deflection. The range can be extended and at the same time the equivalent resistance reduced by connecting a resistor Rsh (shunt resistance) of low resistance in parallel with the meter terminal as shown in figure.
 
 
An expression will now be derived for extending the range off an ammeter for the measurement of a current I greater than Imf.
I = Ish + Imf and VT = Rsh Ish = Rm Imf
Where I = desired range of new ammeter.
Thus,
𝑅𝑠ℎ = 𝐼𝑚𝑓 𝐼𝑠ℎ 𝑅𝑚
𝑅𝑠ℎ = [ 𝐼𝑚𝑓 𝐼 − 𝐼𝑚𝑓] 𝑅𝑚
𝑅𝑠ℎ = [ 0.002 5 − 0.002] 100
𝑅𝑠ℎ = [ 0.002 4.998] 100
𝑅𝑠ℎ = [0.0004002]100
𝑅𝑠ℎ = 0.04002 Ω
With Rsh = 0.04002 Ω in parallel with the original meter, the result is a low resistance Ammeter with the desired range of 5 A.
 
3(b) Range of voltmeter can be extended, and at the same time the equivalent resistance increased. This is done by connecting a resistor of large resistance Rs, called a multiplier resistor, in series with the moving coil as is shown in Figure.
 
 
The current through the meter must be 2 mA when 10V is applied to the terminals to obtain a full-scale deflection. From above Figure, and with Im= Imf,
VT = (Rm + Rs) Imf
From which
𝑅𝑠 = [ 𝑉𝑇 𝐼𝑚𝑓] − 𝑅m
𝑅𝑠 = [ 10 0.002] − 100
𝑅𝑠 = 5000 − 100
𝑅𝑠 = 4900 Ω
Thus, a multiplier resistor of approximately 4.9 kΩ increases the resistance of the voltmeter and extends the range to 10V maximum
 
 

References:


1. https://www.physics.uoguelph.ca/tutorials/ohm/Q.ohm.intro.series.html
2. https://farside.ph.utexas.edu/teaching/302l/lectures/node58.html
3. https://www.electronics-tutorials.ws/resistor/res_5.html
4. https://epb.apogee.net/foe/fcspcom.asp
5. https://www.physicsclassroom.com/class/circuits/Lesson-4/Combination-Circuits
6. https://www.facstaff.bucknell.edu/mastascu/elessonshtml/Resist/Resist2.html
7. https://www.allaboutcircuits.com/vol_1/chpt_7/2.html
8. https://physics.stackexchange.com/questions/81976/power-dissipated-in-seriesversus-parallel
9. https://www.allaboutcircuits.com/vol_1/chpt_8/4.html
10. https://www.myodesie.com/index.php/wiki/index/returnEntry/id/2945
11. https://www.allaboutcircuits.com/vol_1/chpt_8/2.html
12. https://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/volmet.html
13. https://www.engineersedge.com/instrumentation/electrical_meters_measurement/
simple_voltmeter.htm
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