16.2 – Electrical Power | Physics with Mr Shone

Electrical Power

Definition: Electrical Power
Power is energy supplied per unit time.

SI unit for power is Watt (W)

In equation form:

$P = I V$

Where:

$P$ = power of the electrical appliance: watts (W)
$I$ = electric current flowing through appliance: amperes (A)
$V$ = potential difference across appliance: volts (V)

by substituting in V = I R, we can get the following variations:

$P = I^2 R$

$P=\frac{V^2}{R}$

Example 1 (P = IV)
What is the power dissipated in resistor R? $P = I V$ $P$ = 1.5 × 12 $P$ = 18 W

Example 1 (P = IV)

What is the power dissipated in resistor R?

$P = I V$

$P$ = 1.5 × 12

$P$ = 18 W

Example 2 (P = V²/R)
What is the power dissipated in the 10 Ω resistor? $P=\frac{V^2}{R}$ $P$ = 8² / 10 $P$ = 6.4 W

Example 2 (P = V²/R)

What is the power dissipated in the 10 Ω resistor?

$P=\frac{V^2}{R}$

$P$ = 8² / 10

$P$ = 6.4 W

Example 3 (P = I² R)
What is the power dissipated in the 15 Ω resistor? $P = I^2 R$ $P$ = 2² × 15 $P$ = 60 W

Example 3 (P = I² R)

What is the power dissipated in the 15 Ω resistor?

$P = I^2 R$

$P$ = 2² × 15

$P$ = 60 W

Example 4
What is the power dissipated in the 6 Ω and 12 Ω resistors connected in parallel? P.d. across each resistor will be 12 V as they are both in parallel with the cell. For 12 Ω resistor: $P=\frac{V^2}{R}$ $P$ = 12² / 12 $P$ = 12 W For 6 Ω resistor: $P=\frac{V^2}{R}$ $P$ = 12² / 6 $P$ = 24 W

Example 4

What is the power dissipated in the 6 Ω and 12 Ω resistors connected in parallel?

P.d. across each resistor will be 12 V as they are both in parallel with the cell.

For 12 Ω resistor:

$P=\frac{V^2}{R}$

$P$ = 12² / 12

$P$ = 12 W

For 6 Ω resistor:

$P=\frac{V^2}{R}$

$P$ = 12² / 6

$P$ = 24 W

For resistors in parallel it is actually the resistor with lower resistance which will use the most power.

As the electrical energy is converted into heat here the 6 Ω resistor will get hotter than the 12 Ω resistor.

The battery is supplying 36 W to the circuit.

Example 5
What is the power dissipated in the 6 Ω and 12 Ω resistors connected in series? As both resistors are in series there will be the same current flowing through them which will be given by: I = V / R I = 12 / 18 = ⅔ A For 12 Ω resistor: $P = I^2 R$ $P$ = ⅔² × 12 $P$ = 5.3 W For 6 Ω resistor: $P = I^2 R$ $P$ = ⅔² × 6 $P$ = 2.7 W

Example 5

What is the power dissipated in the 6 Ω and 12 Ω resistors connected in series?

As both resistors are in series there will be the same current flowing through them which will be given by:

I = V / R

I = 12 / 18 = ⅔ A

For 12 Ω resistor:

$P = I^2 R$

$P$ = ⅔² × 12

$P$ = 5.3 W

For 6 Ω resistor:

$P = I^2 R$

$P$ = ⅔² × 6

$P$ = 2.7 W

For resistors in series it is the resistor with larger resistance which will use the most power.

As the electrical energy is converted into heat here the 12 Ω resistor will get hotter than the 6 Ω resistor.

The battery is supplying 8.0 W to the circuit.

Example 6
A bulb is marked as “60 W, 240 V”. (a) What is the current drawn from the bulb when used correctly? $P = I V$ 60 = $I$ × 240 $I$ = 0.25 A (b) What is the resistance of the 60 W bulb? $P=\frac{V^2}{R}$ 60 = 240² / $R$ $P$ = 960 Ω (c) The bulb is plugged in in another country whose mains electricity is delivered at 120 V. What is the power that the bulb produces in this instance? Assuming that there is no change in the resistance of the bulb, then: $P=\frac{V^2}{R}$ $P$ = 120² / 960 $P$ = 15 W

Example 6

A bulb is marked as “60 W, 240 V”.

(a) What is the current drawn from the bulb when used correctly?

$P = I V$

60 = $I$ × 240

$I$ = 0.25 A

(b) What is the resistance of the 60 W bulb?

$P=\frac{V^2}{R}$

60 = 240² / $R$

$P$ = 960 Ω

(c) The bulb is plugged in in another country whose mains electricity is delivered at 120 V. What is the power that the bulb produces in this instance?

Assuming that there is no change in the resistance of the bulb, then:

$P=\frac{V^2}{R}$

$P$ = 120² / 960

$P$ = 15 W

Note:

The bulb does not use 60 W when not connected to a 240 V supply.
As the voltage is halved, the current will also be halved, and the power will then one quarter of the original value (60 W ÷ 4 = 15 W)

Electrical Power

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