Voltage and Current sources (Ch. 2.1)
Voltage and Current sources (Ch. 2.1)
(a) An ideal voltage source is a circuit element that maintains a prescribed
voltage across its terminals regardless of the current flowing in those
terminals.
(b) An ideal current source is a circuit element that maintains a
prescribed current through its terminals regardless of the voltage across
those terminals.
Batteries – Alkaline, Silver-oxide, Lead-acid etc.
In any battery, an electrochemical reaction moves ions
from one pole to the other. The actual metals and
electrolytes used control the voltage of the battery -each different reaction has a characteristic voltage.
In an alkaline battery, the anode (negative terminal) is
made of zinc powder (which allows more surface area
for increased rate of reaction therefore increased
electron flow) and the cathode (positive terminal) is
composed of manganese dioxide.
Solar Cells
“Solar revolution" is the idea that one day we will all
use free electricity from the sun.
On a bright, sunny day, the sun shines approximately
1,000 watts of energy per square meter of the planet's surface
1967 - Soyuz 1
was the first manned spacecraft to be powered by solar cells
The spacecraft crashed during its return to Earth. One of
the reasons: the left solar panel deployment failure
Hydroelectric Power Station
Hydroelectricity is electricity generated by hydropower, i.e., the production of power
through use of the gravitational force of falling or flowing water. It is the most widely
used form of renewable energy.
Once a hydroelectric complex is constructed, the project produces no direct waste.
Wind turbine
Wind is a renewable energy because it uses the wind's energy. It is
also a non-polluting energy because it does not produce any green
house effect gases.
Internal resistance of a battery (of a voltage supply)
+
-
+
9V
9V
battery
A
RL
Battery is typically considered
as an example of ideal voltage
source
B
Simple circuit representing the actual battery.
According to this circuit, the voltage across ANY
resistor connected to it, VL = 9V. This is NOT TRUE.
The current flowing in any circuit connected to the
battery also flows THROUGH the battery (through the
electrolyte and the electrodes forming the battery).
Both electrolyte and electrodes have resistance.
This is called “internal” resistance.
Internal resistance of a battery (of a voltage supply)
A
+
-
+
9V
9V
battery
Ri
RL
B
More accurate equivalent circuit of the battery (and many
other voltage sources). Ri is the internal resistance of the
battery. Gray rectangle represents the battery.
VB = 9 V is the “open-circuit” voltage of the battery.
The voltage across the load resistance, VL = I ×RL;
The current in this circuit, I = VB/(Ri + RL);
VL =VB
The voltage across the load:
Example: RL = 100 Ω; Ri = 20 Ω;
The voltage across the load: VL = 9V *100/120 = 7.5 V
RL
<VB
Ri + RL
Internal resistance of a battery (of a voltage supply)
A
+
-
+
9V
9V
battery
Ri
RL
B
RL
VL = VB
Ri + RL
VL= 0.9VB
VL(RL)
VL= VB/2
RL = Ri
RL = 10*Ri
RL
Problem 1
A 9V battery has an internal resistance Ri = 10 Ohm. The battery is
supplying the power to the fire alarm sensor that has an equivalent
resistance of 50 Ohm.
What is the voltage across the sensor (in Volts)?
0
of
5
120
Timed response
A 9V battery has an internal resistance Ri = 10 Ohm. The battery is
supplying the power to the fire alarm sensor that has an equivalent
resistance of 50 Ohm.
What is the voltage across the sensor?
Solution
VB = 9V
REQ = Ri + RL = 10 Ohm + 50 Ohm = 60 Ohm
I = VB / REQ = 9 V / 60 Ohm = 0.15 A
VL = I * RL = 0.15 A * 50 Ohm = 7. 5 V
Problem 2
A 9V battery has an internal resistance Ri = 10 Ohm. The battery is
supplying the power to the fire alarm sensor that has an equivalent
resistance of 50 Ohm.
What is the current flowing through the sensor (in Amperes)?
0
of
5
120
Timed response
Problem 3
A 1.5V battery has an internal resistance Ri = 20 Ohm.
Two batteries are connected in series to power up the MP3 player with
the equivalent resistance of 410 Ohm.
What is the voltage across the MP3 player (in Volts)?
0
of
5
120
Timed response
A 1.5V battery has an internal resistance Ri = 20 Ohm.
Two batteries are connected in series to power up the MP3 player with
the equivalent resistance of 410 Ohm.
What is the voltage across the MP3 player?
Solution
VB = 2 * 1.5 V = 3 V
REQ = 2*Ri + RL = 2*20 Ohm + 410 Ohm = 450 Ohm
VL = VB * RL /REQ = 3 V* 410 Ohm /450 Ohm = 2.73 V
Problem 4
To determine the internal resistance of a battery the test is set up as
follows.
1. The open circuit voltage has been measured and found to be 12.4 V.
2. The voltage at the battery terminals was measured with the 20- Ohm
load connected and found to be 11.9 V.
What is the internal resistance of the battery (in Ohms)?
0
of
5
120
Timed response
1. The open circuit voltage has been measured and found to be 12.4 V.
2. The voltage at the battery terminals was measured with the 20- Ohm
load connected and found to be 11.9 V.
What is the internal resistance of the battery?
Solution
VB = 12.4 V
RL = 20 Ohm
REQ = Ri + RL - unknown variable.
VL = VB * RL /REQ
REQ = VB*RL/VL = 12.4 V*20 Ohm/11.9 V = 20.84 Ohm;
Ri = REQ – RL = 20.84 Ohm – 20 Ohm = 0.84 Ohm
Power delivered to the load
A
+
-
+
9V
9V
battery
Ri
RL
B
Assuming an ideal battery, the power delivered to the load
resistor RL, PL = I ×VL = (VB/RL) ×VB = VB2 /RL.
According to this, reducing the load resistance one can obtain
infinitely high power from the battery.
Using more realistic equivalent circuit of the battery, the power in the load,
PL = I ×VL, where
RL
VB
VL =VB
I=
and
Hence, the power
Ri + RL
RL
2
PL =VB
( Ri
+ RL )
Ri + RL
2
Power delivered to the load
A
+
-
+
9V
9V
battery
PL =VB2
Ri
RL
B
RL
( Ri + RL )
2
Example: RL = 100 Ω;
For ideal battery, the power in the load would be:
PL = VB2 /RL = 92 V2/100 Ω = 0.81 W
For real battery with the internal resistance Ri = 20 Ω,
PL = 92 V2 *[100/(100+20)2] = 81*100/14400 = 0.56 W
Problem 5
The vacuum cleaner rated power is 1500 W with the nominal voltage
V=120 V.
When the vacuum is plugged into the outlet, the actual voltage across
the cleaner was 112 V.
What was the actual power consumed by the vacuum (in Watts)?
0
of
5
120
Timed response
The vacuum cleaner rated power is 1500 W with the nominal voltage
V=120 V.
When the vacuum is plugged into the outlet, the actual voltage across the
cleaner was 112 V.
What was the actual power consumed by the vacuum?
Solution
The nominal current of the vacuum cleaner
IN = PN/VN = 1500 W / 120 V = 12. 5 A
The resistance of the unit
RU = VN /IN = 9.6 Ohm
The actual current through the unit:
I = V/RU = 112 V/9.6 Ohm = 11.6 A
The actual power:
P = V * I = 112 V * 11.6 A = 1306 W
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