Using IB symbols - sketch a series circuit including battery lamp heating element wires with 1 switch The current direction real & conventional . Show the current measurement with an ammeter symbol. Show the measurement of p.d . around bulb and battery with the voltmeter.
By conservation of charge, & because there is only one route, the current (I) is the same everywhere.I1 = I2 = I3 …1, 2, 3 represent circuit components.
Resistance route, the current (I) is the same everywhere. on a series circuit.As more resistors are added Rtot increases.The total/ equilvalent resistance is: R1 + R2+ R3 = Req.
R route, the current (I) is the same everywhere.1
Apply Ohm’s Law to individual components on circuit
The p.d. or V across each resistor is always: V1 = IR1. V2 = IR2. V3 = IR3. Add each device voltage to find the VT.The eq(total) resistance is the sum.
R1+ R2 + R3 = Req.
Since there is only one path for the charges to follow if one conductor (resistor) is disconnected, the circuit is broken. The current flow stops.
As you add more cells, the total voltage adds.
Since each resistor is connected the energy of the charges decreases. across the voltage, the voltage is equal everywhere, and equal to the battery voltage.V1 = V2 =V3 = Vtot
Current (I), when the charges reach a junction, they divide. The total current in the circuit = S individual currents in each branch. Itot = I1 + I2 + I3 …Since I = V, & all components have same V, thenR Itot = V + V + V R1 R2 R3.
As you add resistors, the total equivalent resistance goes down, the total current goes up. The battery or source provides unlimited current!!
Since parallel circuits offer more than one path for the charge to flow, individual parts can be disconnected and charge will still flow through other branches.
Since the voltage is equal on each branch, adding more branches does not reduce the energy each branch receives. Add more bulbs, the others stay bright!
Voltmeters branches does not reduce the energy each branch receives. Add more bulbs, the others stay bright! measure potential difference & are connected in parallel around the component to be measured.
Ex 2: A 9V battery is connected in branches does not reduce the energy each branch receives. Add more bulbs, the others stay bright!parallel to 2 bulbs: 4W, & 2W. A) Sketch the diagram with the proper symbols. Show real and conventional current flow direction. B) Sketch a voltmeter reading the voltage at the 2 W resistor and 2 ammeters: one reading the total current and one reading the current at the 4 W resistor.
B) Find the equivalent or total resistance on the circuit. branches does not reduce the energy each branch receives. Add more bulbs, the others stay bright! C) Find the total current in the circuit. D) What is the voltage in each branch? E) Find the current in each branch.
F) Add the currents from each branch together. How do they relate to the total current?G) Now add a 3 W bulb to the circuit. Recalculate the equivalent resistance.H) Recalculate the current.
Ex: 3 bulbs are in parallel connected to 6V. R relate to the total current?1 = 2W, R2 = 3 W, R3 = 4W.
Which equation is best to calculate power?
Find the power in each bulb.What is relative brightness?What if the 2W is removed? What happens to the brightness of the remaining bulbs?
Hwk relate to the total current? Kerrread 137 – 141 do pg147 #1, 4, 6, 8, 9, 12, 13, 15, 22.
Power relate to the total current?
Ex: 2 bulbs in series connected to 6V source. R relate to the total current?1 = 2W, R2 = 4 W.Find the power in each.What is relative brightness?What if the 2W is removed? What is the power in the remaining bulb?
For parallel I would use P = V relate to the total current?2/R
Ammeters relate to the total current? measure current so circuit current must flow through meter.
Connect meter in series to measure current.
Fuses Application of – appliances are rated for the power they can safely dissipate.That implies a certain current & voltage (power). Fuses should be chosen to have a current rating a bit higher than the one for which the resistor is designed.
If a 60 W bulb is connected to a 120 V source, the current is: P = VI I = P/V. I = 60 J/s = 0.5 C/s .5A. 120 J/CFuse should be ~ .6-1 A.