## SERIES AND PARALLEL CIRCUITS CHAPTER 23

SERIES AND PARALLEL CIRCUITS CHAPTER 23 xx SERIES CIRCUITS • One pathway through the circuit. • All current flows through each resistor. –Current t...
Author: Edith Williams
SERIES AND PARALLEL CIRCUITS CHAPTER 23

xx

SERIES CIRCUITS • One pathway through the circuit. • All current flows through each resistor. –Current through each resistor is the same.

120 V R1

R3

R2

• If the circuit opens (bulb burns out), other circuit items (bulbs) stop working. • If another circuit item (bulb) is added, the other circuit items do not “work” as well. (Bulbs dim) Voltage gets divided between the circuit items.

Resistance in Series The total resistance is the sum of the individual resistances.

RT = R1 + R2 + R3 + ...

Voltage in Series The total voltage drop around the circuit is equal to the voltage on the power supply. Each circuit item “uses up” part of the total voltage.

VT = VR1 + VR2 + VR3 + ...

Current in Series The current through a series circuit is constant. The current through each resistor is the same. Ohm’s Law on the IT = VT entire circuit. RT

IT = IR1 = IR2 = IR3 = ...

(IC#1) A 5 ohm and a 10 ohm resistor are placed in series across a 45 V battery. A. Draw the circuit. Include all meters and a switch.

45 V R1 = 5 Ω R2 = 10 Ω

B. Find RT, IT, VR1, and VR2.

RT = R1 + R2 + R3 + ... RT = R1 + R2 RT = 5 Ω + 10 Ω IT = VT RT

RT = 15 Ω

IT = 45 V 15 Ω

IT = 3 A

VR1 = IR1 ( R1 ) I is constant in series. VR1 = 3 A ( 5 Ω ) = 15 V VR2 = 3 A ( 10 Ω ) = 30 V V adds in series.

45 V

(IC#2) What resistance would have to be added in series with a 10 ohm hair dryer plugged into a 120 V outlet to reduce the total current to 4 amps?

IT = VT RT RT = 120 V 4A

RT = VT IT RT = 30 Ω

RT = R1 + R? 30 Ω = 10 Ω + R?

R? = 20 Ω

Parallel Circuits Multiple pathways (branches) through the circuit. Current splits up and some flows through each branch. The sum of all the flows equals the total current.

R1 R2 120 V

• If the circuit opens (bulb burns out), other circuit items (bulbs) keep working. • If another circuit item (bulb) is added, the other circuit items continue to “work” as well. (Bulbs do not dim) Voltage drop across each branch stays the same.

Current in Parallel The current through a simple parallel circuit is additive. The current through each resistor adds to the total current. IT = IR1 + IR2 + IR3 + ...

Voltage in Parallel The voltage drop across each branch in a simple circuit is the same as the power supply.

VT = VR1 = VR2 = VR3 = ...

Resistance in Parallel IT = IR1 + IR2 + IR3 + ... VT = VR1 + VR2 + VR3 + ... RT R1 R2 R3 VT = VR1 = VR2 = VR3 = ...

1 = 1 + 1 + 1 + ... RT R1 R2 R3 Placing resistors in parallel always decreases the total resistance of the circuit. RT in parallel will always be less than the smallest resistance in the parallel section.

This presents a problem with parallel circuits. The more branches in a parallel circuit, the less resistance. As R ↓ , I ↑ As I ↑ , the amount of heat generated increases. This creates a fire hazard.

(IC#3) Three resistors (60 ohm, 20 ohm, 10 ohm) are connected in parallel across a 90 V battery. A. Draw the circuit. Include all meters and a switch.

R1 = 60 Ω R2 = 20 Ω 90 V

R3 = 10 Ω

B. Find RT, IT, VR1, VR2, VR3. IR1, IR2, and IR3. 1 = 1 + 1 + 1 + ... RT R1 R2 R3 1 = 1 + 1 + 1 = RT 60 Ω 20 Ω 10 Ω 1 = 10 RT 60Ω

RT = 6 Ω

IT = VT RT

IT = 90 V 6Ω

IT = 15 A VT = VR1 = VR2 = VR3 = ... VR1 = VR2 = VR3 = 90 V

IR1 = VR1 R1

IR2 = VR2 R2

IR1 = 90 V 60 Ω

IR2 = 90 V 20 Ω

IR1 = 1.5 A

IR2 = 4.5 A

IR3 = VR3 R3 IR3

IR3 = 90 V 10 Ω = 9.0 A

IT = IR1 + IR2 + IR3 + ... IT = 1.5 A + 4.5 A + 9A IT = 15A

I in parallel adds

(IC#4) What resistance would have to be added in parallel with a 40 ohm hair dryer to reduce the total resistance to 8 ohms?

1 = 1 + 1 RT R1 R2 1 = 1 + 1 X 8Ω 40 Ω 4 = 1 X 40 Ω

X = 10 Ω

Disadvantages of Parallel Circuits • As more circuit items are placed in the circuit, R↓ because there are additional pathways in the circuit. • This causes I↑. • ↑ I causes additional heat.

• Additional heat can cause overheating and a fire. –This is called overloading the circuit.

Protection from Overloads • Fuses are placed in a circuit to prevent overloads. • Fuses open a circuit if the current becomes too great.

Thin metal filament Filament has a low melting point. If T↑ to a predetermined ( ? Amps) level because of an ↑ I , the filament melts and stops the flow of current.

• Circuit breakers are also placed in a circuit to prevent overloads. • These devices are automatic switches that open a circuit if I↑ past a predetermined level. • Circuit breakers can be reset once the overload situation is resolved.

10 A 90 V

3A

18 A

5A

Fuse 16 A

Will this circuit open?

Yes

Short Circuits Normal Circuit I← I→

I Shorted out Current is rerouted. circuit. Since the resistance in a wire is low, the I↑. I↑ causes a large increase in heat. Sparks and/or fire can occur.

The End