A current of 6 A is flowing through an electrical appliance with a potential difference of 220 V - NSC Technical Sciences - Question 9 - 2022 - Paper 1

To calculate the power dissipation in the electrical appliance, we can use the formula:

$P = VI$

where:

• $P$ is the power in watts,
• $V$ is the potential difference in volts (220 V), and
• $I$ is the current in amperes (6 A).

Substituting the values:

$P = 220 imes 6$

Thus,

$P = 1320 ext{ W}$

Therefore, the power dissipated in the electrical appliance is 1320 W or 1.32 kW.

We know that in a series circuit, the total voltage is the sum of the voltage drops across each resistor. The total voltage provided by the battery is 1.5 V, and the voltage drop across R1 can be calculated using Ohm's Law:

$V = IR$

For R1:

• Current ($I$) = 0.5 A,
• Resistance ($R1$) = 2 Ω.

Thus,

$V_{R1} = I imes R1 = 0.5 imes 2 = 1 ext{ V}$

Now, we can find the voltage drop across R2:

$V_{R2} = V_{total} - V_{R1} = 1.5 - 1 = 0.5 ext{ V}$

Using this voltage drop to find the resistance of R2:

Using Ohm’s Law again:

$V_{R2} = I imes R2$

Substituting for $R2$:

$0.5 = 0.5 imes R2$

So,

$R2 = 1 ext{ Ω}$