Applying electrical equations

Charge, Current, and Ohm's Law Calculations

When calculating electrical quantities, time must always be converted into seconds. Failing to convert minutes to seconds is a common way to lose marks.

Calculate the charge that flows through a lamp when a current of $3.5\text{ A}$ is maintained for $4\text{ minutes}$.

Step 1: Identify and convert values.

• $I = 3.5\text{ A}$

• $t = 4\text{ minutes} = 4 \times 60 = 240\text{ s}$

Step 2: State the formula and substitute.

• $Q = I \times t$

• $Q = 3.5 \times 240$

Step 3: Calculate the final result with units.

• Result: $840\text{ C}$

A heating element has a resistance of $15\ \Omega$. Calculate the potential difference required to drive a current of $0.6\text{ A}$ through it.

Step 1: State the formula.

• $V = I \times R$

Step 2: Substitute and calculate.

• $V = 0.6 \times 15$

Step 3: Final result with units.

• Result: $9.0\text{ V}$

Electrical Work Done and Energy Transfer

In electrical circuits, Electrical Work Done is exactly equivalent to energy transferred. When charge moves through a potential difference, energy is shifted from the power supply to the components.

A battery moves $8.0\text{ C}$ of charge through a potential difference of $6.0\text{ V}$. Calculate the electrical work done.

Step 1: State the formula.

• $E = Q \times V$

Step 2: Substitute and calculate.

• $E = 8.0 \times 6.0$

Step 3: Final result with units.

• Result: $48\text{ J}$

An electric motor has a power rating of $45\text{ W}$. How much energy is transferred if it operates for $20\text{ seconds}$?

Step 1: State the formula.

• $E = P \times t$

Step 2: Substitute and calculate.

• $E = 45 \times 20$

Step 3: Final result with units.

• Result: $900\text{ J}$

Electrical Power and Power Dissipation

Electrical Power is the rate at which energy is transferred, measured in Watts (W). It can be calculated using the potential difference and current ($P = IV$).

Alternatively, when studying the heating effect of a resistor, we use the equation $P = I^2R$. This specifically calculates the power dissipation—the rate at which electrical energy is transferred to the thermal store of the component.

Calculate the electrical power of a bulb that draws a current of $2.5\text{ A}$ from a $12\text{ V}$ power supply.

Step 1: State the formula.

• $P = I \times V$

Step 2: Substitute and calculate.

• $P = 2.5 \times 12$

Step 3: Final result with units.

• Result: $30\text{ W}$

A resistor has a resistance of $150\ \Omega$ and a current of $3.0\text{ A}$ flowing through it. Calculate the power dissipation.

Step 1: State the formula.

• $P = I^2 \times R$

Step 2: Substitute the values, ensuring the current is squared.

• $P = (3.0)^2 \times 150$

• $P = 9.0 \times 150$

Step 3: Final result with units.

• Result: $1350\text{ W}$

Solving Circuits with Resistors in Series

A Series Circuit is a circuit with only one single loop, meaning there is only one path for the charge to follow. Consequently, the current is identical at all points in the circuit.

If you place multiple resistors in a series circuit, they combine to create a Total Resistance (or Equivalent Resistance). You calculate this by simply adding the individual resistances together:

A circuit contains a $16\text{ V}$ battery connected in series to a $5.0\ \Omega$ resistor and a $3.0\ \Omega$ resistor. Calculate the current flowing through the circuit.

Step 1: Find the total equivalent resistance.

• $R_{total} = R_1 + R_2$

• $R_{total} = 5.0 + 3.0 = 8.0\ \Omega$

Step 2: State Ohm's Law formula for current.

• $I = \frac{V}{R_{total}}$

Step 3: Substitute and calculate.

• $I = \frac{16}{8.0}$

Step 4: Final result with units.

• Result: $2.0\text{ A}$