Electromotive force and potential difference

e.m.f. = Work Done / Charge

e.m.f. = 12 J / 4 C

e.m.f. = 3 V

The electromotive force of the battery is 3 Volts. It represents the energy provided by the battery per unit charge passing through it.

The unit missing is volts (V).

Therefore, the correct measurement is 1.5 V. E.m.f is always measured in volts.

A potential difference of 6.0V across the lightbulb means that for every 1 Coulomb of charge that passes through the bulb, 6.0 Joules of electrical energy is transferred into light and heat energy. The charge loses 6.0 J of energy as it moves through the bulb.

The label tells you that the potential difference (p.d.) across the terminals of the battery is 3.0 volts.

Explanation: This means the battery provides 3.0 Joules of electrical energy per Coulomb of charge that passes through it.

1. Select an appropriate range on the voltmeter. If the expected potential difference is unknown, start with the highest range and decrease until a reading is obtained without exceeding the scale.
2. Connect the voltmeter *in parallel* with the resistor. This means connecting one terminal of the voltmeter to each end of the resistor.
3. Ensure the positive terminal of the voltmeter is connected to the point in the circuit with the higher potential (usually the side connected to the positive terminal of the power supply).
4. Read the potential difference from the voltmeter's scale or digital display. The voltmeter displays the potential difference.

You should select the 15V range.

Reasoning: The 3V range is too low as the reading (12V) would exceed the scale and potentially damage the meter. The 30V range could be used, but it would result in a less precise reading as the needle would deflect less, making it more difficult to read accurately. Selecting 15V gives an appropriate reading on the scale.

Formula: e.m.f. = V + Ir

e.m.f. = 5.0 V + (2.0 A * 0.5 Ω)

e.m.f. = 5.0 V + 1.0 V

e.m.f. = 6.0 V

Explanation: The e.m.f. is the total energy supplied per unit charge, which is the sum of the terminal potential difference (V) and the potential drop across the internal resistance (Ir).

When a battery supplies current, some energy is dissipated as heat due to the internal resistance of the battery. This energy loss results in a voltage drop (Ir) within the battery itself. Therefore, the terminal potential difference (V), which is the voltage measured across the battery's terminals, is equal to the e.m.f. minus the voltage drop due to the internal resistance (V = e.m.f. - Ir). Since Ir is always positive when a current is flowing, the terminal potential difference is always less than the e.m.f.

Formula: V = W / Q, where V = potential difference, W = energy, and Q = charge.
Rearranging: Q = W / V
Calculation: Q = 1200 J / 6.0 V = 200 C
Answer: The charge that flows through the circuit is 200 Coulombs (C).

A potential difference of 9.0 V across a component means that 9.0 Joules (J) of energy are transferred per Coulomb (C) of charge that passes through the component. The charge loses 9 J of electrical energy to the component, which is then converted to other forms of energy.