Electric Circuits



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 1 1 1 1

------- = ------ + ------ + ------ + … + ------

REQ 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.

Vs ∙ R1

V1 = -------------

(R1 + R2)

Vs ∙ R2

V2 = -------------

(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.

I ∙ R2

I1 = -------------

(R1 + R2)

I ∙ R1

I2 = -------------

(R1 + R2)

Circuit Example

[pic]

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.

[pic]

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

[pic]

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

................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download