9.1 Define power in an electric circuit - NSC Technical Sciences - Question 9 - 2021 - Paper 1

To find the resistance of the heater element, we can use the power formula:

P = rac{V^2}{R}

Rearranging this gives:

R = rac{V^2}{P}

Substituting in the values:

• Power, $P = 60 ext{ W}$
• Voltage, $V = 220 ext{ V}$

We calculate:

R = rac{(220)^2}{60} = rac{48400}{60} = 806.67 ext{ Ω}

Thus, the resistance of the heater element is approximately 806.67 Ω.

First, we need to determine the total resistance in the circuit, which consists of a 3 Ω resistor and a 5 Ω resistor in series:

$R_{total} = R_1 + R_2 = 3 ext{ Ω} + 5 ext{ Ω} = 8 ext{ Ω}$

Next, we find the total voltage supplied by the cells. Since there are two cells each with a potential difference of 1.5 V:

$V_{total} = 1.5 ext{ V} + 1.5 ext{ V} = 3 ext{ V}$

Using Ohm's law, we can calculate the total current in the circuit:

I = rac{V_{total}}{R_{total}} = rac{3 ext{ V}}{8 ext{ Ω}} = 0.375 ext{ A}

Now, we can find the potential difference across the 5 Ω resistor using:

$V = I imes R$

Substituting in our values:

$V_{5Ω} = 0.375 ext{ A} imes 5 ext{ Ω} = 1.875 ext{ V}$

Thus, the potential difference across the 5 Ω resistor is 1.875 V.