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ELECTRIC CIRCUITS
Ohm’s Law
V = I ∙ R (Some books use E = I ∙ R)
where:
V = voltage (volts or V)
I = current (amperes or amps or A)
R = resistance (ohms or Ω)
Equivalent Resistance
Series Resistance:
Given resistors R1, R2, R3, …, RN; connected in series.
The equivalent resistance is given by the formula:
REQ = R1 + R2 + R3 + … + RN
Parallel Resistance:
Given resistors R1, R2, R3, …, RN; connected in parallel.
The equivalent resistance is given by the formula:
1 / REQ = 1 / R1 + 1 / R2 + 1 / R3 + … + 1 / RN
1
------- =
REQ
1
1
1
1
------ + ------ + ------ + … + -----R1
R2
R3
RN
For the “special” case of two resistors (R1 and R 2)
connected in parallel, the formula becomes:
REQ = (R1 ∙ R2) / (R1 + R2)
R1 ∙ R2
REQ = ------------R1 + R2
Voltage Law
The voltage changes around any closed loop in an electric
circuit must sum to zero.
Voltage Divider Rule
For a circuit with two resistors, R1 and R2, in series, the
voltage drop across each resistor equals the resistance times
the total voltage, divided by the sum of the two resistors.
V1 =
Vs ∙ R1
------------(R1 + R2)
V2 =
Vs ∙ R2
------------(R1 + R2)
Current Law
The electric current which flows into any junction in an electric
circuit is equal to the current which flows out.
Current Divider Rule
For two parallel resistors, R 1 and R2, the current through the
branch equals the resistance of the opposite branch times the
input current divided by the sum of the two resistors.
I1 =
I ∙ R2
------------(R1 + R2)
I2 =
I ∙ R1
------------(R1 + R2)
Circuit Example
If R1 = 24 Ω, R2 = 24 Ω, R3 = 12 Ω, R4 = 12 Ω, R5 = 12 Ω and
Vs = 12 Volts, determine the circuit equivalent resistance
(REQ) and the circuit current (I). Also calculate the branch
currents through R2, R3, R4, and R5, and the voltage drops
across R1, R2, R3, R4 and R5.
Solution
Calculate R45 (Series Resistance)
R45 = R4 + R5 = 12 + 12
R45 = 24 Ω
Calculate R345 (Parallel Resistance)
1/R345 = 1/R3 + 1/R45 = 1/12 + 1/24 = 2/24 + 1/24 = 3/24
Therefore, R345 = 8 Ω
Calculate R2345 (Parallel Resistance)
1/R2345 = 1/R2 + 1/R345 = 1/24 + 1/8 = 1/12 + 3/24 = 4/24
Therefore, R2345 = 6 Ω
Calculate REQ (Series Resistance)
REQ = R1 + R2345 = 24 Ω + 6 Ω
REQ = 30 Ω
Calculated the circuit current. Use Ohm’s Law V = I ∙ R
12 V = I ∙ 30 Ω
Therefore, I = 12/30
I = 0.4 A
Calculate the voltage drop across R1
V1 = I ∙ R1 = 0.4 A ∙ 24 Ω
V1 = 9.6 V
Calculate the voltage drops across R2 and R3
V2 = V3 = 12 V - 9.6 V
V2 = V3 = 2.4 V
Calculate the current through R2
I2 = V2 / R2 = 2.4 V / 24 Ω
I2 = 0.1 A
Calculate the current through R3
I3 = V3 / R3 = 2.4 V / 12 Ω
I2 = 0.2 A
Calculate the current through R4 and R5
The voltage drop across R45 = V2 = V3 = 2.4 V
Therefore I4 = I 5 = V45 / R45 = 2.4 V / 24 Ω
I4 = I5 = 0.1 A
Finally, calculate the voltage drops across R4 and R5
V4 = I4 ∙ R4 = 0.1 A ∙ 12 Ω
V4 = 1.2 V
V5 = I5 ∙ R5 = 0.1 A ∙ 12 Ω
V5 = 1.2 V
PROJECTILE MOTION
Definitions
Projectile - Any body in freefall that has a horizontal aspect to
its motion.
Trajectory - The curve that describes the motion of a body in
space.
Motion in Two Dimensions
The x and y components of motion are completely separable.
At any time, a projectiles velocity can be divided up in the
following manner.
Projectile Motion Equations
Acceleration
Ax = 0
Ay = -g
Velocity
Vx = V(cos )
Vy = V(sin ) + Ayt
= V(sin ) - gt
Position
Px = Px0 + V(cos )t
Py = Py0 + V(sin )t + 0.5Ayt2
= Py0 + V(sin )t - 0.5gt2
Javelin Example
A javelin thrower releases a javelin 1.5 meters from the ground
at an angle of 48. If the initial velocity of the javelin is 25 m/s,
what is the distance the javelin travels.
Components of the initial velocity:
V(cos ) = 25(cos(48)) = 16.73 m/s
V(sin ) = 25(sin(48)) = 18.58 m/s
Initial and final position values:
Px0 = 0 m;
Py0 = 1.5 m;
Py = 0 m;
Use the following equation to determine t:
Py = Py0 + V(sin )t - 0.5gt2
0 = 1.5 + 18.58t - 4.9t2
Using the quadratic equation to solve for t:
t = 3.871 s
We can now solve for Px:
Px = Px0 + V(cos )t = 0 + 16.73(3.871) = 64.76 m
Find the maximum height of the javelin.
Remembering that the vertical velocity is zero when
the javelin reaches its highest point.
Vy = V(sin ) - gt
0 = 18.58 - 9.8t
t = 1.896 s
We can now solve for Pymax
Pymax
= Py0 + V(sin )t - 0.5gt2
= 1.5 + 18.58(1.896) - 4.9(1.896)2
= 19.11 m