A student investigates resistors connected in series in an electrical circuit - Edexcel - GCSE Physics - Question 6 - 2020 - Paper 1

The voltage across the 22Ω resistor is 0.9 V.

Using the formula $V = I \times R$, we can rearrange it to find resistance:

$R = \frac{V}{I} = \frac{2.1V}{0.041A} \approx 51.22 \Omega$

This indicates that the resistance of resistor X is approximately 51 ohms, which rounds to about 50 ohms.

Using the power formula:

$P = V \times I = 2.1V \times 0.041A = 0.0861W$

This rounds to approximately 0.086 W.

The total resistance in a series circuit is the sum of individual resistances:

$R_{total} = R_{22\Omega} + R_{X} = 22\Omega + 51\Omega = 73\Omega.$

Therefore, the overall resistance is 73 ohms.

Using the energy formula:

$E = I \times V \times t$

where $I = 0.041A$, $V = 3.0V$, and $t = 120s$ (2 minutes), we have:

$E = 0.041A \times 3.0V \times 120s = 14.76 J$

This rounds to about 15 J.