Question

An electric heater of power $1470W$ has a resistance of $30\Omega$. Calculate the magnitude of

1. the current
2. the p.d. at its end.

Answer 1

Step 1 - Given data and formula

Power $P=1470W$

Resistance $R=30\Omega$

Formula for power is given by the expression

$P=I^2\times R$

Where $I$ is the current flowing through the circuit.

Step 2 - Finding the formula for current

Rearrange the above formula.

$$\begin{gathered} P=I^2\times R \\ {I}^2=\frac{P}{R} \\ I=\sqrt{\frac{P}{R}} \end{gathered}$$

Step 3 - (a) Finding the current

Substitute values into the formula for $I$.

$$\begin{gathered} I=\sqrt[]{\frac{P}{R}} \\ I=\sqrt{\frac{1470W}{30\Omega }} \\ I=7A \end{gathered}$$

Step 4 - (b) Finding potential difference at the ends of the heater

From the expression for Ohm's law, the potential difference $V$ can be calculated as

$$\begin{gathered} V=I\times R \\ V=(7A)\times (30\Omega ) \\ V=210V \end{gathered}$$

Final answer - a. The current is $7A$.

b. The p.d at the end of the heater is $210V$.