Microsoft Word - Assignment_4_V7_2DME.docx Introductory Physics Assignment 3 – V7 CQUniversity Australia 1 Introductory Physics Assignment 3 This test covers material from the modules: ELEC, MAGN and...

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Answer To: Microsoft Word - Assignment_4_V7_2DME.docx Introductory Physics Assignment 3 – V7 CQUniversity...

Priyanka answered on Sep 12 2021
160 Votes
ELEC-Electricity
1. R1=35 Ω and R2 =65 Ω resistors are connected in parallel.
a) Electrical circuit
+
-
24V 35ᾨ 65ᾨ
IT
I1 I2

b) Equivalent resistance=Req
Parallel combination of two resistors (R1,R2)can be
replaced with a resistor(Req)
1
Req
=
1
R1
+
1
R2

Simplified further as:
Req=
?1 ∗ ?2
R1+R2

Outing values gives
Req =
35∗65
35+65
=22.75 Ω
c) Potential difference across each resistor.
V1 =potential difference across R1
V2 =potential difference across R2
R1 and R2 are connected in parallel so voltage (Potential difference) will be same also
both resistors are connected to power supply of24 V.
i.e. V1= V2=24V
d) I1 = current through R1
I2 = current through R2
Using Ohm’s law
I1 =
V1
R1
and I2=
V2
R2

Putting values
I1=
24
35
=0.6857A
I2=
24
65
=0.3692A
d) P1 = power dissipated in R1= I12* R1
P2 = power dissipated in R2= I2
2* R2
Putting values
P1= (0.6857)
2 *35=16.457 Watt
P2= (0.3692)
2 *65=8.8615 Watt
MAGN (MAGNETISM)
1. Two parallel wires separated by 16 cm are carrying currents of 14 A (left) and 22A (right) in
the same direction (up the page). Draw a diagram to represent this and the magnetic field
directions. Determine the magnetic field at position 3 cm to the right of the 22A wire. Calculate
the distance where the magnetic field would be zero
16 cm
Left Right
14 A
22 A
Magnetic field lines

16 cm
14A
22A
r1=3cm
P
r2 =19 cm

Magnetic field at point P = sum of magnetic field due to individual wires on point P=B
Magnetic field lines are drawn
according to right hand thumb rule
Current up to the paper is
represented with (●)
I1=22A
I2=14A
Magnetic field at point P due to 22 A wire= B1
Magnetic field at point P due to 14 A wire=B2
B=B1+B2
? =
?0 ?1
2 ∗ ? ∗ ?1
+
?0 ?2
2 ∗ ? ∗ ?2

Putting the values ?0 : p
?0 ?2
2∗?∗?2
ermeability =4 ∗ ? ∗ 10−7 H/m
? =
4 ∗ ? ∗ 10−7 ∗ 22
2 ∗ ? ∗ 0.03
+
4 ∗ ? ∗ 10−7 ∗ 14
2 ∗ ? ∗ 0.19

= 2 ∗ 10−7 (
22
0.03
+
14
0.19
) =16.14*10-5 T
Point at which magnetic field becomes 0
i.e. 0=B1+B2
B1=-B2
?0 ?1
2 ∗ ? ∗ ?1
= −
?0...
SOLUTION.PDF

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