Potential difference and power

A water heater of resistance `60 Ω` runs from a mains supply of `230` V. It can raise the temperature of a tank of water from `20`^oC to `45`^oC in 20 minutes. Calculate:
a. The charge that passes through the heater.
b. The energy dissipated by the heater.

a. `R = V / I`

Which leads to `I = V / R = 230 / 60 = "3.83 A"`.

`Q = I xxt = 3.83 xx 20 xx 60 = "4600 C"`.

b. `V = W / Q`

Which leads to `W = V xxQ = 230 xx 4600 = "1058000 J" approx "1.1 MJ"`.

Electrons in a particular television tube are accelerated by a potential difference of 20 kV between the filament and the screen. The charge of the electron is `-1.6 × 10^(-19)` C. Calculate the gain in kinetic energy of each electron.

Since `V=W/Q`, then `W=VxxQ`. The energy is transferred to the electron increasing its kinetic energy.

Thus, `"kinetic energy gained" = Vxx Q = 20 xx 10^3 xx 1.60 xx 10^-19= 3.2 xx 10^-15 " J"`
Don’t forget to convert the 20 kV into volts.

A torch bulb is rated 2.2 V, 0.25 A. Calculate:
a. The charge passing through the bulb in one second.
b. The energy transferred by the passage of each coulomb of charge.

a. `Q = I × t= 0.25 × 1 = "0.25 C"`
b. `E = V × Q= 2.2 × 1 = "2.2 J"` 

The capacity of storage batteries is rated in ampere-hours (Ah). An 80 Ah battery can supply a current of 80 A for 1 hour, or 40 A for 2 hours, and so on. Calculate the total energy, in J, stored in a 12 V, 80 Ah car battery.

Convert ampere-hours (Ah) to coulombs (C): 
`Q = "80 Ah" = 80 × 3600 = "288000 C"`
Energy stored in a battery: `E = Q × V = 288000 × 12 = "3456000 J"`

An electric kettle is rated at 2.2 kW, 240 V. The supply voltage is reduced from 240 V to 230 V. Calculate the new power of the kettle.

The formula relating power and voltage (when resistance is constant):

`P ∝ V^2 ⇒ P_2 = P_1 × (V_2 / V_1)^2` 

`P_2 = 2.2 × (230 / 240)^2 = 2.2 × (0.9583)^2 = 2.2 × 0.918 ≈ "2.02 kW"`

What is the definition of the potential difference (p.d.) across a component?
A. The energy transferred per unit charge.
B. The energy transferred per unit current.
C. The power transferred per unit charge.
D. The power transferred per unit current.

Answer: A

A. Correct: Potential difference (V) is defined as: `V = W/Q`. This tells us how much energy is transferred or converted per coulomb of charge passing through the component.
B. Incorrect: This describes not p.d., but rather a time-dependent concept.
C. Incorrect: Power is energy per unit time, so dividing power by charge is not dimensionally related to potential difference.
D. Incorrect: This gives potential difference multiplied by time, again unrelated to the actual definition of voltage.

Two lamps are connected in series to a 250 V power supply. One lamp is rated 240 V, 60 W and the other is rated 10 V, 2.5 W. Which statement most accurately describes what happens?
A. Both lamps light at less than their normal brightness.
B. Both lamps light normally.
C. Only the 60 W lamp lights.
D. The 10 V lamp blows.

Answer: B
One lamp is rated 240 V, 60 W, this means that for the lamp to operate normally, there should be a potential difference of 240 V across it.

Current flowing through this lamp `I=P / V = 60 / 240 = "0.25 A"`

The other lamp is rated 10 V, 2.5 W. This lamp operates normally when there is a potential difference of 10 V across it.

Current flowing through the lamp `I=P / V = 2.5 / 10 = "0.25 A"`

So, for normal operation, the same current of 0.25 A is required for both lamps. Since the lamps are connected in series, the same current would flow. Hence, both lamps light at their normal brightness.

A power cable X has resistance R and carries current `I`.

A second cable Y has resistance 2R and carries current `I/2`

Determine the ratio `("power dissipated in Y") / ("power dissipated in X").

The power dissipated in X is `P = Vxx I=RxxI^2`

The power dissipated in Y is `P=(1/2)^2 xx (2R) = (2 I^2 R) / 4 = (I^2 R) / 2`

Therefore, the ratio: 

`("power dissipated in Y") / ("power dissipated in X") = ((I^2 xxR) / 2)/ (RxxI^2) = 1 / 2`

The amount of energy transferred when 10 C of charge passes through a potential difference of 20 V is the same as the energy needed to raise a 2 kg mass through a distance `x`. Take gravitational field strength to be 10 N.kg^-1. What is the value of `x`?

The amount of energy transferred: `Vxx Q = 20 xx 10 = "200 J"`
Gravitational potential energy: `=mxxgxxh`
Where `h` is the vertical distance moved by the mass, `h =x`.
Equate this with the energy transferred by the charge: `mxx gxx x = "200 J"`

`x = 200 / (2 xx 10) = "10 m"`

Two lightbulbs, X and Y, are in parallel and are attached to a cell which has a potential difference of 75 V. Lightbulb X has a resistance of 350 Ω and Lightbulb Y has a resistance of 500 Ω. The power of Lightbulbs X and Y are `P_X` and `P_Y` respectively.

What is the value of the ratio `(P_X)/(P_Y)`

A. `1/2`

B. `7/10`

C. `10/7`

D. `2`

Answer: C

The power of the lightbulbs can be found using `P = V^2 / R`

The potential difference for each light bulb will be the same as they are in parallel, therefore `P ∝ 1 / R`

The power of Lightbulbs X and Y are `P_X ∝ 1 / 350` and `P_Y ∝ 1 / 500`

The value of the ratio of  `(P_X)/(P_Y)` is `500/350=10/7`