Presentation on theme: "Physics Mrs. Coyle. Kirchhoff’s Rules Series Circuits Equivalent Resistance Voltage Drop Across Resistors Brightness of Bulbs in a Series Circuit."— Presentation transcript:
Physics Mrs. 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
1 st Rule: (Junction Theorem): At a junction (node), current in= current out 2 nd 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 = V 1 + V 2 + V 3
If the battery’s voltage is 12V and the voltage across R 1 is 5 V, and across R 2 is 4V, find the voltage across R 3. Answer: 3V
V = V 1 + V 2 + V 3 Using Ohm’s Law: IR eq = IR 1 +IR 2 +IR 3 Equivalent resistance R eq = R 1 + R 2 + R 3
If the battery’s voltage is 12V and R 1 = 1 Ω R 2 = 2 Ω R 3 = 3 Ω Find the equivalent resistance. Find the current. Find the voltage across each resistor. Answer: 6 Ω, 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=I 2 R P=V 2 / 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