Calculate The Effective Resistance | Assessment Answer

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 â„¦ + 4 â„¦ + 9 â„¦

= 15 â„¦

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 â„¦

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 â„¦

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 â„¦

Now, Ra, R3 & R4 are in series, so their equivalent effective resistance is

Reff = Ra + R3 + R4

= 1.33 + 4 + 5

Reff = 10.33 â„¦

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 â„¦

Now, Ra & R3 are in series, so their equivalent effective resistance is

Reff = Ra + R3

= 0.833 + 4

Reff = 4.833 â„¦

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 â„¦

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 â„¦

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 â„¦

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,

ð‘…ð‘ â„Ž = ð¼ð‘šð‘“ ð¼ð‘ â„Ž ð‘…ð‘š

ð‘…ð‘ â„Ž = [ ð¼ð‘šð‘“ ð¼ − ð¼ð‘šð‘“] ð‘…ð‘š

ð‘…ð‘ â„Ž = [ 0.002 5 − 0.002] 100

ð‘…ð‘ â„Ž = [ 0.002 4.998] 100

ð‘…ð‘ â„Ž = [0.0004002]100

ð‘…ð‘ â„Ž = 0.04002 â„¦

With Rsh = 0.04002 â„¦ 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

ð‘…ð‘ = [ ð‘‰ð‘‡ ð¼ð‘šð‘“] − ð‘…m

ð‘…ð‘ = [ 10 0.002] − 100

ð‘…ð‘ = 5000 − 100

ð‘…ð‘ = 4900 â„¦

Thus, a multiplier resistor of approximately 4.9 kâ„¦ 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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