Basic EE Week 4 Lesson

THE ELECTRIC CIRCUIT

A.
LEARNING OUTCOMES
B.
After completing this
unit, you are expected to:

1. describe a basic electric circuit.

2. discuss the function of a fuse or circuit breaker.
3. discuss the causes of overcurrent.
4. determine the rating of fuse or circuit breaker that fits to
a given circuit.
5. identify closed, open, and short circuit .
6. measure voltage, current, and resistance using both
analog and digital multimeters.
7. apply Ohm’s law to determine voltage, current, and
resistance in electric circuits.
8. apply Kirchhoff’s Current Law (KCL) and Kirchhoff’s
Voltage Law (KVL) in a given circuit.
 Important Terms

electric circuit Closed circuit

energy source Open circuit
Conductors Digital multimeter
Insulators Analog multimeter
load Ohm’s law
Control device overcurrent
Protection
device Short circuit
Fuse overloading
Circuit breaker SPST
SPDT

4.1 The Electric Circuit

Electric Circuit is an interconnection of electrical components in which there is at

least one path for current flow.

Many complete electric circuits contain six parts:

1. An energy source to provide the voltage needed to force current (electrons) through the
circuit.
2. Conductors through which the current can travel.
3. Insulators to confine the current to the desired paths (conductors, resistors, etc.)
4. A load to control the amount of current and convert the electric energy taken from the
energy source.
5. A control device, often a switch, to start and stop the flow of current.
6. A protection device to interrupt the circuit in case of a circuit malfunction.

The first four of the above six parts are essential parts. All complete circuits use them. The
control device (item 5) is occasionally omitted. Protection devices (item 6) are often omitted from
circuits. A complete electric circuit has an uninterrupted path for current (electrons) to flow from
the negative terminal of the energy source through the load and control device to the positive
terminal of the energy source. An example of a circuit is shown in Figure 4.1 and Figure 4.2.
 Figure 4.1. A circuit consisting of a battery (voltage source), a switch, a buzzer (load), and
wires used to interconnect the components.

4.2 Protection Devices

Circuit Breakers Fuses

A fuse is inserted into a circuit to protect the device / circuit from receiving too much current
when shorted.

A circuit breaker is the same function. If a surge of current is evident through a line, the circuit
breaker "breaks" the line, opening the flow of current.

A fuse breaks the circuit only once, then, has to be replaced.

A circuit breaker is a "switch" that closes and opens the line. A circuit breaker is usually
STRONGER (tolerates more current flow than a fuse.)

4.2.1 Causes of Overcurrent

a. Short Circuit
b. Overloading
Example 4.1 You are an electrician on the job. The electrical blueprint shows that eight 500-W
lamps are to be installed on the same circuit. The circuit voltage is 230 V and is protected by a 20-
A circuit breaker. Is a 20-A circuit large enough to carry this load?

Solution:

IT

20 A

1 Vpk
2X6U2
12
X3X2V
230 V 12 412V1
X4 1kHz
3V 12V
12
1VV
X1
X5
X8 U1
12 VX7
0°

20 A

total power
Total current (IT ) =
total voltage

8 x 500 W
Total current (IT ) = = 17.39 A
230 V

Since the total current in the circuit is 17.39 A, therefore, a 20-A circuit breaker or fuse is large
enough to carry the load.

Example 4. 2 If the load in example 4.1 is a continuous load, is the 20-A circuit breaker can carry
the load. Note that a continuous-use circuit can be loaded only 80% of its rating.

Solution:
The 20-A circuit breaker can carry only 16 A (80% of 20 A) since the load is
continuous. Since the current in the circuit is 17.39 A, the circuit breaker cannot carry this load.
4.3 Circuit Components and Symbols

Device Symbol

Lamp (light bulb)

X1
12 V

12V1
V
Cell or battery
 Fuse
U1
0.5_AMP

Circuit Breaker

Single-Pole-Single KeyOpen
=J1
Space KeyClosed
=J2
Space
Throw (SPST)
Switch

Single-Pole-Double
Throw (SPDT)
Switch

Push Button

Relay
 Conductor

Resistor R

1kΩ

L
Inductor (coil)
1mH

Capacitor C

1µF

DC Motor

Ground
 switch

battery
Key J1
=V
V1 Space
bATTERY
16
12 X1
V

(a) (b)

Figure 4.2. (a) A simple electric circuit. (b) Schematic diagram for the circuit in (a)

4.4 Closed, Open and Short Circuit

Closed Circuit

A closed circuit is a circuit in which the current has a complete path.

Key J1
=V
V1 Space
battery
bATTERY
switch
16
12 X1
V

Open Circuit
When the current path is broken so that current cannot flow, the circuit is called an open
circuit.

Key J1
=V
V1 Space
battery
bATTERY
switch
16
12 X1
V
Short Circuit

A short circuit is a zero or abnormally low resistance between two points. It is usually an
inadvertent condition.

Key = Space KeyJ1

= Space
J1
X1battery
bATTERY
switch
12 V
V1 12 VX1 12 VV1
16 V

Difference Between Open Circuit and Short Circuit

V=  V= 0

I=0 I=

R= R=0

Open Circuit Short Circuit

In an open circuit, it is possible to have a voltage depending one the network; the current is zero
because it is an open circuit so definitely current cannot flow through it; the resistance is infinite.

In a closed circuit, the voltage is equal to zero; current is possible to flow depending on the
network; resistance is equal to zero. But in real circuit, there is a certain amount of resistance in
the short circuit line but that resistance is nearly equal to zero.

4.5 Basic Circuit Measurement

Three common electrical quantities are voltage, current, and resistance. These quantities are
measured by voltmeter, ammeter, and ohmmeter. They are usually put into one instrument
called multimeter, multitester, or VOm (voltmeter, ohmmeter, and milliammeter). Two types of
multitester are digital and analog.
 + - + -
00.000 A 00.000 V

XMM1
Ammeter VoltmeterOhmmeter

Meter Symbols
How to Measure Current with an Ammeter

E R
60ohm
12V

Circuit in which the current is to be measured

(a) Open the circuit either between the resistor and the negative terminal or between the
resistor and the positive terminal of source.

E R
60ohm
12V

(b) Install the ammeter with polarity as shown (negative to negative – positive to positive)
 + -
0.200 A

E R
60ohm
12V

.
How to Measure Voltage with a Voltmeter

To measure voltage, connect the voltmeter across the component for which the voltage is to be
found. Such a connection is a parallel connection. The negative terminal of the meter must be
connected to the negative side of the circuit and the positive terminal of the meter to the positive
side of the circuit.

R + -
E
12.000 V
60ohm
12V

How to Measure Resistance with an Ohmmeter

To measure resistance, connect the ohmmeter across the resistor. The resistor must be removed
or disconnected from the circuit.

XMM1

E R
60ohm
R
12V 60ohm

(a) Disconnect the resistor from the (b) Measure the resistance.
circuit to avoid damage to the (polarity is not important.)
meter and/or incorrect
measurement.
4.6 The Analog Multitester

How to read the measurements in an analog multitester.

1. Select what quantity is to be measured by using the rotary switch.

Reminder: The rotary selector switch below selects what quantity is to be

measured. If you are measuring a voltage of 8 V it must be pointed to the 10-V
range. THE RANGE SHOULD BE HIGHER THAN THE EXPECTED VALUE OF
VOLTAGE, CURRENT, OR RESISTANCE.

2. Connect the meter as discussed in 4.4

3. The measured value of the electrical quantity can be read from the needle. Using the scales
below (Figure 4.3) , if you are measuring DC voltage and selector switch placed at 50-V
range, the DC voltage will read 22 V.

Figure 4.3

Example 4.3 Using Figure 4.3, find the correct reading if the selector is placed at the following:

DC 10V range
DC 50V range
DC 25mA range
AC 10V range

Solution:

DC 10V range: 4.4V (read 0-10 scale directly)

DC 50V range: 22V (read 0-50 scale directly)
DC 25mA range: 11mA (read 0-250 and divide by 10)
AC 10V range: 4.45V (use the red scale, reading 0-10)
 Objective Test No. 6

ELECTRIC CIRCUIT MEASUREMENT

1. An instrument that measures the voltage or electrical pressure in a circuit.

A. megger
B. ammeter
C. voltmeter
D. galvanometer

2. A multimeter consists of a
A. ammeter and ohmmeter
B. voltmeter and ammeter
C. voltmeter and ohmmeter
D. voltmeter, ammeter, and ohmmeter

3. How is a voltmeter connected in a circuit?

A. Connect in series across the load
B. Connect in shunt across the load
C. Connect in open circuit with the load
D. Connect in short circuit across the load

4. Which of the following is an integrating instrument?

A. ammeter
B. ohmmeter
C. voltmeter
D. wattmeter

5. An ammeter is connected ________.

A. across the load
B. in series with the load
C. in series-parallel across the load

6. Electrical measurement used to measure electrical power

A. wattmeter
B. galvanometer
C. clamp ammeter
D. kilowatt-hour meter

7. Instrument used to measure electrical energy.

A. wattmeter
B. galvanometer
C. clamp ammeter
D. kilowatt-hour meter
4.7 Ohm’s Law

For a simple DC circuit, German physicist George Simon Ohm has observed another
relationship related to the resistance of an object. He noted that for a fixed load in a circuit at a
constant temperature, when voltage is increased the current reading across the load also is
increased. Consequently, as the voltage is lowered across the load, current reading is also lowered.
Therefore, he arrived at the conclusion that current is directly proportional to voltage. And
when the ratio between voltage and current was computed, it has been found out that it is
approximately equal to the resistance of the load. Thus, Ohm’s Law states that: “The ratio between
voltage and current is always constant and is equal to the resistance of the load. (At a constant
temperature!)

E R
60ohm
12V

V
R = I

where:
I = current in amperes
V = voltage in volts
R = resistance in ohms, 

Other Ohm’s Law Formulas

V
V = IR I = R

The Conductance
Conductance is the reciprocal of resistance. It is the property of a conductor or resistor to
allow current flow.

Symbol: G

Unit: Siemens (S)

The formulas are

I I
G = I = GV V =
V G

Limitation of Ohm’s Law

Ohm’s law does not apply under the following condition:

1. Electrolytes where enormous gases are produced on either electrode.

2. Non-linear resistors like vacuum radio valve, semi-conductors, gas-filled tubes, etc.
3. Arc lamps
4. Metals which get heated up due to flow of current through.
5. Appliances like metal rectifiers, crystal detectors, etc. in which operation depends on the
direction of current.
Example 4.4 For the circuit shown below, find the current through the 60- resistor.

E R
60ohm
12V

Solution:
V 12 V
I = = = 0.2 A
R 60 

Example 4.5 The difference of potential between the terminals of an electric heater is 110 V when
there is a current of 8 A in the heater. What current will be maintained on the heater if the
difference of potential is increased to 180 V?

Solution:
V1 110 V
R = = = 13.75 
I1 8A

V2 180 V
I2 = = = 13.09 A
R 13.75 

Example 4.6 A 6-V battery is connected for 3 hours to a rheostat and a current of 147 mA is noted.
(a) What is the resistance of the rheostat? (b) What charge is take form the battery?

Solution:
V
(a) R = = 40.8 
I

3600 sec
(b) Q = It = (147 mC/s)(3 hours x 1 hr )= 1.59 x 103
Assessment No. 8

OHM’S LAW

Name: Last, First, MI Score: ____________

1. Find the unknown in circuit below.

I = _______ A
+ -
1.000 A

V R +
12.000 V
12ohm -
12V
2. Show the placement of an ammeter and voltmeter to measure the current and
voltage across s R1 and R2.

R1
12ohm

12V

R2
12ohm

3. The current in a 5- resistor increases linearly from zero to 10 A in 2 ms. At t = 2 ms the current
is again zero, and it increases linearly to 10 A at t = 4 ms. This pattern repeats each 2 ms. Sketch
the corresponding v.

4. A certain electrical device has an unknown resistance. You have available a 12-V battery and
an ammeter. How would you determine the value of the unknown resistance? Draw the
necessary connections.
 Problem Set No. 7

OHM’S LAW

1. For the circuit shown below, find the current through the 60- resistor.

A. 0.002 A
B. 0.02 A E R
60ohm
C. 0.2 A 12V

D. 2A

2. The difference of potential between the terminals of an electric heater is 110 V when there is a
current of 8 A in the heater. What current will be maintained on the heater if the difference of
potential is increased to 180 V?

A. 13.09 A
B. 0.02 A
C. 0.2 A
D. 2A

3. A 6-V battery is connected for 3 hours to a rheostat and a current of 147 mA is noted. (a) What
is the resistance of the rheostat? (b) What charge is taken from the battery?

A. 40.8 ohms, 1,587.6 C

B. 32.8 ohms, 1,587.6 C
C. 40.8 ohms, 1,556.6 C
D. 34.8 ohms, 1,487.6 C

4. Find the value of the current measured by the ammeter.

R
30Ω

U1
E -
36 V 0.000 A DC 1e-009 W
+

A. 1.2 A
B. 0.02 A
C. 0.2 A
D. 2A

5. Find R in the circuit below.

A. 12 Ω
B. 13 Ω
 C. 14 Ω
D. 16 Ω

+ U1
I -64.000 V DC 10M W
R -
4A
16Ω

6. What is the potential difference across a 15 -  resistor when a current of 6.5 A passes through
it?

A. 67.5 V
B. 77.5 V
C. 87.5 V
D. 97.5 V

Objective Test No. 7

OHM’S LAW

1. When using Ohm’s Law, E divided by I would solve for

A. watts
B. voltage
C. amperage
D. resistance

2. The condition of Ohm’s law is that

A. the temperature should vary
B. ratio V/I should be constant
C. current should be proportional to voltage
D. the temperature should remain constant

3. If the resistance of the circuit is doubled while the applied voltage is held constant. The
current will ____________________.
A. be twice as much
B. remain the same
C. increase by half as much
D. decrease to half as much

4. A circuit has a resistance of 8 ohms. If a voltmeter connected across its terminals reads 10 V,
how much current is flowing through the circuit?
A. 0.80 A
 B. 1.25 A
C. 1.5 A
D. 2.10 A

5. If the potential across a circuit is 40 V and the current is 5,000 mA, what is the equivalent
resistance of the circuit?
A. 8 
B. 80 
C. 800 
D. 800 k

6. The difference of potential between the terminals of an electric heater is 120 V when they are
at a current of 8 A. What current will be maintained in the heater if the potential difference is
increased to 220 V?
A. 4.4 A
B. 12 A
C. 14.7 A
D. 40 A

7. When using Ohm’s law “IR” would solve for ___________.

A. voltage
B. amperage
C. resistance
D. electrical power

8. A water heater takes 2.5 A at 230 V. What is its hot resistance?

A. 74 ohms
B. 82 ohms
C. 92 ohms
D. 100 ohms
4.8 Kirchhoff’s Law
.
.
Important Terminologies

Node – a point in which two or more components have a common connection.

Path- if no node was encountered more than once, then the sets of nodes and elements that we
have passed through is defined as a path.

Closed Path or Loop- If the node at which we started is the same as the node on which we ended,
then the path, is by definition, a closed path or loop.

Branch - is a single path in a network, composed of one simple element and the node at each end
of that element.

R3
1kohm
R1 R2
I 1kohm 1kohm
1A
2
R4
1kohm

(a)

R3
1kohm
R1 R2
I 1kohm 1kohm
1A
2
R4
1kohm

(b)

(a) A circuit containing three nodes and five branches. (b) Node 1 is redrawn to look like two
nodes; it is still one node.

4.8.1 Kirchhoff’s Voltage Law (KVL)

The algebraic sum of all voltages in a circuit taken around a closed path is zero.
Sign Convention a

V
12V
 Path b – a : Potential Rise = + V
Path a – b: Potential Drop = -V

+ a

R
1ohm VR

I
_ b

Path b – a: Potential Rise = +VR

Path a – b: Potential Drop = - VR

4.8.2 Kirchhoff’s Current Law (KCL)

The algebraic sum of all currents entering and leaving a node is zero.

Convention: I2

I3

I1
I4

Current entering a node: + sign

Current leaving a node: -sign

I1 – I2 – I3 + I4 = 0

Example 4.7 Find V1

- +
14.000 V

V2
a b a
R
2V 30ohm
V1 V3
12V 24V

V4

e d
4V
Solution:

V1 + V2 + VR – V3 – V4 = 0

V1 = − V2 − VR + V3 + V4

V1 = − 2 − 14 + 24 + 4

V1 = 12 V

Example 4.8 Find V2

Solution:
V1 + V2 + VR − V4 = 0

V2 = − V1 − VR + V4

V2 = − 12 − (−10) + 4
V2 = 2 V

Example 4.9 Find the current in the circuit.

2VA

+ V30 - 𝑖
-
VA

+
Solution:

Let the current i be in the clockwise direction.

By applying KVL in a clockwise direction.

120 − V30 − 2 VA + VA = 0
By Ohm’s law
V30 = 30i
VA = −15i

so that 120 − 30i − 2(−15i) − 15i = 0

i=8A

Example 4.10 Find I3

R1
a

15ohm
R2
I1 36ohm I2

3A 5A
I3

b
Solution: At node a

I1 + I2 − I 3 = 0

I3 = I1 + I2
I3 = 3 + 5
I3 = 8 A

Example 4.11 Find i3 and i6.

i3 i6

0.9i3

Solution: By applying KCL at either node.

0.9i3 + 2 − i3 − i6 = 0
 v6 3i3 1
where i6 = = = 2 i3 . Substituting to the equation, then
6 6

1
0.9i3 + 2 − i3 − i =0
2 3

i3 = 3.33 A

Example 4.12 Calculate the current flowing in each branch of the circuit shown

R2
V1 V3
20ohm
3V 6V

R1 R3
V2
10ohm 20ohm
4.5V

Solution:

Step 1: Assign branch current direction on a node and form KCL equation. (NOTE: Do not be
over-conscious about the direction. If you wrongly assigned it, it will just yield negative answer
but the magnitude will be the same.)

R2
V1 V3
20ohm
3V 6V

I2
R1 R3
V2
10ohm 20ohm
I 1 4.5V I3

KCL equation:

I1 + I2 + I3 = 0

Step 2: Assign loop-current direction on each visible loop and form your KVL equations. (Again,
direction is arbitrary)

KVL equation:

For loop 1 (L1):

R2
V1 V3
20ohm
10I1 + 3 − 20I2 + 4.5 = 03V 6V

10I1 − 20I2 = − 7.5 L1 I2

R1 L2 R3
V2
10ohm 20ohm
I1 4.5V I3
For loop 2 (L2):

4.5 + 6 − 20I2 + 30I3 = 0

−20I2 + 30I3 = −10.5

Step 3: Solve for the magnitude of the branch currents (I1, I2 and I3) using any method.

I1 + I2 + I3 = 0 equation 1
10I1 − 20I2 = − 7.5 equation 2
−20I2 + 30I3 = −10.5 equation 3

The answers are: I1 = −0.15 A

I2 = 0.3 A

I3 = 0.15 A
.
Example 4.13 Solve for the currents in all the resistors of the circuit shown below.

R1
50ohm

E1 R3

12V
100ohm
E2
R2 3V
200ohm

Solution:

Step 1: Assign branch-current direction and form KCL equation.

R1
50ohm

I1
E1 R3

12V I3 1kohm
E2
R2 3V
50ohm
I2

KCL equation:
−I1 + I2 − I3 = 0

Step 2: Assign Loop-Current Direction and form KVL equations.

R1
50ohm

Loop 1: I1
E1 R3

12 + 50I1 + 200I2 = 0
12V I3 100ohm
E2
L1 R2 3V
50I1 + 200I2 = −12 200ohm
I2
L2
Loop 2:
−3 – 200I2 − 100I3 = 0
−200I2 − 100I3 = 3

Step 3: Solve by matrix and determinants

−I1 + I2 − I3 = 0 equation 1

50I1 + 200I2 = −12 equation 2

−200I2 − 100I3 = 3 equation 3

I1 = 0.086 A; I2 = 0.0385 A; I3 = −0.0475 A

Assignment No. 7 Kirchhoff’s Law

Name: 𝐹𝑖𝑟𝑠𝑡, 𝑀𝐼, 𝐿𝑎𝑠𝑡 Date: 𝑀𝑜𝑛𝑡ℎ 𝐷𝑎𝑦, 𝑌𝑒𝑎𝑟

Course: 𝑇𝑦𝑝𝑒 𝐶𝑜𝑢𝑟𝑠𝑒 𝐶𝑜𝑑𝑒 ℎ𝑒𝑟𝑒 Section: 𝑇𝑦𝑝𝑒 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 ℎ𝑒𝑟𝑒

1. Find the current through each resistor using Kirchhoff’s law.

R1
50ohm

E1 R3

12V
100ohm
E2
R2 3V
200ohm
2. Find the values of I1 and V5.

R1

1ohm

R2
5ohm I2
I1
I1
6A

+
R3 10A
2ohm V5
4A -
R4 1kohm

3ohm

.
3. Find the voltage across the 4-ohm resistor.
 Problem Set No. 8

KIRCHHOFF’S LAW

1. What is the value of V in the circuit shown below?

6A

+ 4V -
+ + + -
-4A 2V 2A 2V 4A -10 A
2V -2 V
- +
- -
-6 A
i -4 A

+ V -
A. 4V
B. 6V
C. 8V
D. 10 V

2. What is the value of the power received by element B?

-7 A

+ 4V -
- - -
+
B 7A
-2 A -2 V -2 V -2 A v
2V i
+ + +

- 7A

- -9 V +
A. -49 W
B. 49 W
C. 94 W
D. -94 W

3. Find I7.
 I2 -2 A

+ -1 V - - 3V +
- -
-1 A I7 I6
2V v6
+ +
I3 I5

+ v3 - - -3 V +
- - +

A. 0.5 A
B. 0.75 A
C. -1 A
D. 1.25 A

4. Find vx.

+
Vx

-
A. 2.4 V
B. 4.2 V 4Vx
C. 4.8 V
D. 5.2 V

5. Find iA , iB and iC.

 iA iB iC

0.1Vx

A. 0.25 A, 1.35 A, 1.5 A

B. 0.75 A, 1.35 A, 1.5 A
C. 0.25 A, 1.35 A, 1.75 A
D. 0.25 A, 1.85 A, 1.5 A

6. What is the value of V7?

I2
+ V7 -

+ 12 V -
+ + - +
I3 2A I6
V1 3A 3V 4V V6
- -
+ -
I4

+ 6V -

A. 11 V
B. 12 V
C. 13 V
D. 14 V

- +
7. Find V1. 14.000 V

V2
a b a
R
A. 22 V 2V 30ohm
V1 V3
B. 23 V 12V 24V

V4

e d
 C. 24 V
D. 25 V

8. Find V2.

A. 1V
B. 1.5 V
C. 1.75 V
D. 2V

9. Find I3.
R1
a

15ohm
A. 8A R2
I1 36ohm I2
B. 9A
3A 5A
C. 10 A I3
D. 12 A
b

10. Calculate the current flowing in each branch of the circuit shown.

R2
V1 V3
20ohm
3V 6V

R1 R3
V2
10ohm 20ohm
4.5V

A. I1 = 112.27 mA, I2 = 317.96 mA, I3 = 205.68 mA

B. I1 = 112.27 mA, I2 = 217.96 mA, I3 = 205.68 mA
C. I1 = 12.27 mA, I2 = 317.96 mA, I3 = 205.68 mA
D. I1 = 112.27 mA, I2 = 317.96 mA, I3 = 105.68 mA

11. Solve for the currents in all the resistors of the circuit shown below.
 R1
50ohm

E1 R3

12V
100ohm
E2
R2 3V
200ohm

A. I1 = 55.62 mA, I2 = 38.54 mA, I3 = 47.07 mA

B. I1 = 85.62 mA, I2 = 18.54 mA, I3 = 47.07 mA
C. I1 = 85.62 mA, I2 = 38.54 mA, I3 = 47.07 mA
D. I1 = 85.62 mA, I2 = 38.54 mA, I3 = 37.07 mA

12. Determine I1, I2, and V2.

R3
6A

I1 5ohm
R1
2 R2 V2
V1 1kohm 6ohm 12V
12V
I2

A. I1 = 1A, I2 = 2 A, V2 = 42 V
B. I1 = 2 A, I2 = 2 A, V2 = 62 V
C. I1 = 2 A, I2 = 2 A, V2 = 42 V
D. I1 = 2 A, I2 = 3A, V2 = 42 V

13. Determine I1 and V5.

R1

1ohm

R2
5ohm I2
I1
I1
6A

+
R3 10A
2ohm V5
4A -
R4 1kohm
 A. I1 = -9 A, V5 = 80 V
B. I1 = -8 A, V5 = 80 V
C. I1 = -8 A, V5 = 90 V
D. I1 = -8 A, V5 = 100 V

14. Find the branch currents in the circuit below.

R1

9.1ohm
10 ohms
V1

16V V3
R3

8ohm
4V

V2

10V
R2

2ohm

A. I1 = 1.09 A, I2 = 159.95 mA, I3 = 2.25 A 2 A

B. I1 = 2.09 A, I2 = 157.95 mA, I3 = 2.25 A 2 A
C. I1 = 2.09 A, I2 = 157.95 mA, I3 = 2.35 A 2 A
D. I1 = 1.09 A, I2 = 197.95 mA, I3 = 2.25 A 2 A
Assessment No. 10: Electric Circuit

Name: Date:
Course: Section: Score:

How Much Have You Learned?

Directions: Solve the crossword puzzle. Use the given clues to arrive at the right answer.
2
1 1 4
3
2

10
8
6
5
7 3

7
5
8 10

DOWN ACROSS
1 current cannot flow 1 current can flow
2 melts in over current 2 short circuit and overloading
3 protection against current 3 equivalent to volt/ampere
4 interconnection of components 4 part of a circuit that consumes energy
5 used to convey current 5 control device
6 V = IR 6 trips
7 causes overcurrent 7 ____________ multimeter
8 energy source 8 equivalent to ampere/volt
9 single-pole, single-throw 10 equivalent to joule/coulomb
10 connected in series to the load

QUESTIONS: (not part of Assessment No. 10)

1. What is an electric circuit?

2. What are the parts of an electric circuit?
3. What is the function of a fuse or circuit breaker?
4. What are the causes of overcurrent?
5. What is the effect of overcurrent?
6. How do you avoid overcurrent?
7. What is a closed circuit? open circuit?
8. How do you connect the voltmeter in measuring voltage?
9. How do you connect the ammeter in measuring current?
10. What is Ohm’s law?
Practical Application No. 3: Electric Circuit

Name: Date:
Course: Section: Score:

1. You are an electrician on the job. The electrical blueprint shows that eight 500-W lamps are to
be installed on the same circuit. The circuit voltage is 277 V and is protected by a 20-A circuit
breaker. Assuming that the load is continuous, is a 20-A circuit large enough to carry this load?

2. You have been sent to a new home. The homeowner reports that sometimes the electric furnace
trips the 240-V, 60-A circuit breaker connected to it. Upon examination, you find that the
furnace contains three 5000-W heating elements designed to turn on in stages. For example,
when the thermostat calls for heat, the first 5000-W unit turns on. After some period of time,
the second unit will turn on, and then, after another time delay, the third unit will turn on. What
do you think the problem is, and so would your recommendation for correcting? Explain your
answer.
3. You are an electrician installing the wiring in a new home. The homeowner desires that a ceiling
fan with light kits be installed in five different rooms. Each fan contains a light kit that can
accommodate four 60-watt lamps. The voltage source is 220 V. Each fan motor draws a current
of 1.8 amperes when operated on high speed. It is assumed that each fan can operate more than
three hours at a time and therefore must be considered a continuous-duty service. The fans are
to be connected to a 15-ampere circuit. Because the devices are continuous duty, the circuit
current must be limited to 80% of the continuous connected load. How many fans can be
connected to a single 15-ampere circuit? How many circuits will be required to supply power
to all five fans?
 4. A homeowner is installing a swimming pool. You have been asked to install a circuit to
operate a 600-watt underwater light and a circuiting pump. The motor nameplate reveals
that the pump has a current draw of 8.5 amperes. The devices are considered continuous
duty. Can the power to operate both of these devices be supplied by a single 20-ampere
circuit? The voltage source is 220 V.