1. 23-1 Simple Circuits

2. Series Circuits
Brightness of lamps is depended upon the ______________ in a circuit
When all current travels in all of the devices, said to be in Series
B/c current travels through all, current is same through out
I = V source
RA + RB
Equivalent Resistance
Sum of all individual resistance

3. Problems
A 10 ohm resistor, 15 ohm resistor, and a 5 ohm resistor are connected in series across a 90 V battery.
A) What is the equivalent resistance in the circuit?
B) What is the current in the circuit?
Known Unknown
R1 = 10 Ω R = ??
R2 = 15 Ω I = ???
R3 = 5 Ω

4. A) What is the equivalent resistance in the circuit?
R = RA + RB + RC
R = 10 Ω + 15 Ω + 5 Ω
R = 30 Ω

5. B) What is the current in the circuit?
I = V source
RA + RB + RC
I = 90 V
10 Ω + 15 Ω + 5 Ω
I = 3 A

6. Voltage Drop Voltage Divider
A Series circuit used to produce a voltage source of desired magnitude from a higher voltage battery
Solve by rearranging the R = V/I
V = IR
R is the equivalent resistance and V is the potential drop

7. Suppose you have a 9-V battery but need a 5-V potential source.
I = V/R = V / RA +RB
VB = IRB
VB = IRB = V x RB
RA +RB
VB = VRB

8. A) What is the current in the circuit?
Two Resistors, 47 Ω and 82 Ω are connected in series across a 45.0 V Battery
A) What is the current in the circuit?
B) What is the voltage drop in each resistor?
C) The 47 Ω resistor is replaced by a 39 Ω resistor. Will the current increase, decrease, or stay the same?
D) What will happen to the voltage drop across the 82 Ω resistor?
Known Unknown
V source = 45.0 V I = ??
RA = 47 Ω VA = ??
RB = 82 Ω VB = ??

9. A) What is the current in the circuit?
Find equivalent resistance
R = RA + RB
I = V source / R = V source / RA +RB
I = V
47 Ω +82 Ω
I = A

10. Find the voltage drop of each resistor
VA = IRA = (0.349)(47 Ω) = 16 V VB = IRB = (0.349)(82 Ω) = 29 V

11. Calculate Current again using RA and determine new voltage drop
I = V source / R = V source / RA +RB
I = V
39 Ω +82 Ω
= 0.372
VB = IRB = (0.372)(82Ω)
= 31 V

12. A 9.0 V battery and two resistors, 400 ohms and 500 ohms, are connected as a voltage divider. What is the voltage across the 500 ohm resistor?

13. Parallel Circuits A circuit in which there are several current paths
Total current is the sum of the currents through each path and the potential difference across each path is the same
Current through path depends on each resistor

14. Parallel Circuits
When resistance is removed, current through other resistors does not change, only the total current through the circuit

15. Parallel Circuits Finding Equivalent resistance I = IA + IB + IC …
So then….
V = V + V + V …
RA RB RC
1/R = (1/RA) + (1/RB) + (1/RC) …

16. Three resistors, 60.0 Ω, 30.0 Ω, and 20.0 Ω, are connected in parallel across a 90.0 V battery
A) Find the current through each branch of the circuit
B) Find the equivalent resistance of the circuit
C) Find the current through the battery

17. 2) Find the equivalent resistance 1/R = (1/RA) + (1/RB) + (1/RC)
= Ω -1
R = 10.0 Ω
3) Find the Total Current
I = V/R
= 90.0 V / 10.0 Ω
= 9.00 A
So we want the total current equal to the individual currents added together

18. Find the current through each branch IA = V / RA IB = V / RB
= 90.0 V / 60.0 Ω = 1.50 A
IB = V / RB
= 90.0 V / 30.0 Ω = 3.00 A
IC = V / RC
= 90.0 V / 20.0 Ω = 4.50 A