Power in electric circuits
φ1
φ2
As the charge moves through the resistor, its energy dissipates.
+
The potential energy of a unit charge in the point (1) is φ1
E, I
The potential energy of a unit charge in the point (2) is φ2
The potential change = φ1 – φ2 = V12
When the charge Q moves from the point (1) to the point (2), the potential
energy change is
∆W = Q* (φ1 – φ2) = Q*V12
The dissipated power is the energy dissipation W per unit time:
P = ∆W/t = Q* (φ1 – φ2 )/t
If the charge Q moves from the point (1) to the point (2) in t seconds,
then the current passing through the conductor is I = Q/t. Therefore,
P = ∆W/t = I*V12
The power dissipated in any component of electric circuit is equal to
the current through the component times the voltage drop across it:
P = I*V
Power in electric circuits
φ1
φ2
The potential change along the current direction,
+
φ1 – φ2 = V12 > 0
E, I
So, for the positive current direction, I >0
-
And hence,
V>0
P = V×I > 0
The power dissipated in any PASSIVE component,
P = I*V > 0
Power in Electric Circuits
1
I
E=24 EV
0
The potential ϕ1 of the point 1 is +24 V with respect to the “ground point” 0 potential
(the battery provides extra energy to mobile charges).
ϕ1 is equal to the energy acquired by a unit charge moving from point 0 to the point 1.
As the positive charge Q moves from point 0 to point 1 its energy increases by
For positive current direction, I >0,
V<0
P=I×V<0
Voltage sources ( e.g. batteries) generate the power (as oppose to dissipate it)
Power units
The energy W is measured in Joules.
From the potential or voltage definition:
1 Joule = 1C × 1V;
The power (mechanical or electrical) is equal to the energy change per unit time,
P = ∆W/t
The power is measured in Watts (W):
1 W = 1 Joule/1s
1 W = (1C × 1V)/1s
Note that Q/t = I, or 1C/1s = 1 A
1W=1A*1V
Example 1.
The current through the resistor R= 4 kOhm is I= 2 A.
What is the power dissipated in the resistor?
Solution
According to Ohm’s law,
V = I*R;
Hence,
P = I*V = I*(I*R) = I2 * R
P = I2 * R = (2A)2 * 4*103 Ohm = 16*103 W = 16 kW
Example 2.
The voltage drop across the resistor R= 4 kOhm, is V= 2 V.
What is the power dissipated in the resistor?
Solution
According to the Ohm’s law,
V = I*R; or I = V/R
Hence,
P = I*V = (V/R)*V = V2 / R
P = V2 / R = (2V)2 /( 4*103 Ohm) = 10-3 W = 1 mW
Quiz problem 1.
R1
5Ω
VB =
12 V
R2
10 Ω
Car mini heater contains two
heating elements as shown.
What is the total power of the
heater (in Watts)?
0
of
5
180
Timed response
Quiz problem 2.
R1
5Ω
VB =
12 V
R2
10 Ω
Car mini heater contains two
heating elements as shown.
What is the power (in Watts)
dissipated in the heating
element R2?
0
of
5
180
Timed response
Quiz problem 3.
VB =
12 V
R1
5Ω
R2
10 Ω
Car mini heater contains two
heating elements as shown.
What is the power (in Watts)
dissipated in the heating
element R2?
0
of
5
180
Timed response
Quiz problem 4.
VB =
12 V
R1
5Ω
R2
10 Ω
Car mini heater contains two
heating elements as shown.
What is the total power of the
heater (in Watts)
0
of
5
180
Timed response
Power ratings
The light bulb is marked as “120V; 60 W”
What is the power of the light bulb when connected to:
(a) 120 V source?
(b) 100 V source?
(c) 12 V source?
When connected to a 120 V voltage source, the dissipated power in this light bulb
is 60 W – this is a so-called “ RATED or NOMINAL POWER”.
However, if the voltage of the power supply is higher or lower than that, the power
won’t be 60 W.
Neither the power, nor the voltage are the intrinsic characteristics of the light bulb.
Only the resistance is. Suppose the resistance of the light bulb is R. When
connected to a voltage source V, the current through the bulb is:
V
I=
R
Hence, the dissipated power is:
Thus if the light bulb is marked as
“120V; 60 W” its resistance is:
V V2
P = V×I =V× =
R R
V V 2 1202
R = V×I =V× =
=
= 240 Ω
R P
60
Power ratings (cont.)
What is the power of the light bulb when connected to:
(b) 100 V source?
The light bulb resistance, R = 240 Ω
The power when connected to the voltage source VS,
P = V*I = V*(V/R) = V2/R
Hence, the dissipated power is:
P = 1002/240 = 41.667 W
What is the power of the light bulb when connected to:
(c) 12 V source?
P = 122/240 = 0.6 W
Quiz problem 5.
The kettle designed for use in Europe (standard household voltage 220 V)
is rated at 1 kW. A tourist brought the kettle in US. What is the actual
power consumed by the kettle when plugged into US outlet (120 V)?
1.
2.
3.
4.
1 kW
1000 W
297.5 W
545.5 W
120
0%
1
0%
2
0%
3
0%
4
Quiz problem 5. Solution
The kettle produced in Europe (standard household voltage 220 V) is rated
for 1 kW.
A tourist brought the kettle in US. What is the actual power consumed by the kettle
when plugged into US outlet (120 V)?
P = V*I = V2/R
The voltage V is different in Europe and in the US, but the kettle
resistance R is the same.
In Europe, PE = VE2 /R; PE= 1 kW as labeled.
In US, PUS= VUS2 /R;
From these, PUS/PE = VUS2/VE2 = 1202/2202 = 0.2975;
PUS= 0.3 * PE= 0.3* 1 kW = 297.5 W.
Power and efficiency
DC adapter consumes 100 mA current when plugged into 120 V voltage outlet.
The output voltage is 12 V and the load current is 750 mA. What is the efficiency
of the adapter?
Consumed power (i.e. input power):
Pin = 120V*0.1 A = 12 W;
Useful power (i.e. output power, or load power):
Pout = 12 V * 0.75 A = 9 W;
The efficiency η = Pout/Pin = 9W/12W = 0.75 = 75%
Electric Energy
Energy = Power × Time;
This equation is equally true for mechanical and electrical energy.
Electric power:
P = V×I
If the current flows through the resistor (or any other load)
during the time “t”, then the dissipated energy is
E = P × t = V×I × t
Energy is measured in Joules. 1 J is the energy dissipated in the load
with the 1W of power during 1 s:
1 J = 1W × 1 s
Energy consumption is also often measured in kW × hours (kWh).
1 kWh is the energy dissipated in the load with 1 kW power during 1
hour.
1 kWh J conversion:
1 kWh = 103 W × 3600 s = 3.6 ×106 J = 3.6 MJ (Mega-Joules)
Example 1: Electric energy
A desk lamp with 100 W light bulb was left ON while the homeowners left for
vacations for 2 weeks.
For how much this desk lamp increases the electricity bill?
The consumed energy = Power x time.
The light bulb Power = 100 W.
Time =2 weeks = 14 days*24 hours/day = 336 hours.
Energy E = 100 W * 336 hours = 33600 W*hours = 33.6 kWh.
The residential electric energy rate is
Rate=0.1 $ / kWh.
The cost of the electricity consumed by the light bulb,
C = Rate* Energy
C = 0.1 $ (kWh)-1 *33.6 kWh ≈ $3.36
Example 2: Electric energy
A kettle produced in Europe (standard household voltage 220 V)
boils 2 liters of water in 10 min.
A tourist brought the kettle in US. How long will it take to boil 2
liters of water with this kettle in US?
Boiling 2 liters of water requires certain amount of energy EB:
EB = P *t
In Europe: EB= P1*t1, where P1 = V12/ Rkettle.
In US, the required energy is the same, but the kettle power is not:
EB= P2*t2, where P2 = V22/ Rkettle.
Equating the required energy in US and Europe, we obtain:
P1*t1 = P2*t2, or V12/ Rkettle* t1 = V22/ Rkettle * t2, or:
V12* t1 = V22 * t2. Hence. t2 = t 1 *(V1/V2)2;
t2 = 10 min. * (220/120)2 = 10 min. * 3.36 = 33.6 min.