PhysicsMrs. Coyle

Kirchhoff’s Rules Series Circuits

Equivalent Resistance Voltage Drop Across Resistors Brightness of Bulbs in a Series Circuit

There is one current path.

All resistors have the same current.

Positive charges are “pumped” by the battery from low to high potential. V>0

When traversing a resistor with the current, there is a decrease in potential. V<0

1st Rule: (Junction Theorem): At a junction (node), current in= current out

2nd Rule: (Loop Theorem): In a closed loop the sum of the voltages is zero.

In a series circuit the total voltage drop across the resistors equals the sum of the individual voltages.

V = V1 + V2 + V3

If the battery’s voltage is 12V and the voltage across R1 is 5 V, and across R2 is 4V, find the voltage across R3 .

Answer: 3V

V = V1 + V2 + V3

Using Ohm’s Law:IReq = IR1+IR2 +IR3

Equivalent resistanceReq = R1 + R2 + R3

If the battery’s voltage is 12V and R1 = 1Ω R2 = 2Ω

R3 = 3Ω Find the equivalent

resistance. Find the current. Find the voltage

across each resistor.

Answer: 6Ω, 2A, 2V, 4V, 6V

The greater the power actually used by a light bulb, the greater the brightness.

Note: the power rating of a light bulb is indicated for a given voltage, at room temperature and the bulb may be in a circuit that does not have that voltage.

P= I V

P=I2 R

P=V2 / R

Find the total resistance.

Find the current. Find the power

dissipated in each lamp. Which light bulb will be

the brightest and why? Find the total power. How does the total

power compare to the powers of the individual bulbs.

Ans: 450Ω, 0.027A, 0.18W, 0.036W, 0.109W, 250 Ω, 0.324W

250Ω

150Ω

50Ω

12 V

Parallel Circuits Equivalent Resistance Brightness of Light Bulb

Combination Circuits

There is more than one current path.

The voltage across the resistors is the same.

http://www1.curriculum.edu.au/sciencepd/energy/images/energy_ill112.gif

I = I1 + I2 + I3

V =V1=V2=V3

Using Ohm’s Law:V/Req= V/R1 +V/R2 + V/R3

Equivalent Resistance:1/Req= 1/R1 +1/R2 + 1/R3

Find the Req , I’s.

How does Req compare with each R?

Ans: 0.55Ω, I= 22A, (12A, 6A, 4A)

=1Ω=3Ω

=2Ω12V

Why should you not plug in too many appliances in the same outlet in a home?

Ans: 11 Ω, 1.8A, V1=9V, V2=11V, I2=1.1A, I3=0.7A

=20V

=10Ω

=15Ω

=5Ω

http://www.eng.cam.ac.uk/DesignOffice/mdp/electric_web/DC/00123.png

Req 1 = 71.4Ω

Req 2 = 127.3Ω

Req = 198.7Ω I=0.12A V1 = 8.6V

V2 = 15.3V