Distinguish between resistance and resistivity - Leaving Cert Physics - Question 8 - 2017

The resistance decreases because the cross-sectional area increases. Specifically, the area $A$ is proportional to the square of the diameter, so if the diameter is increased by a factor of three, the area increases by a factor of:

$A' = \pi \left( \frac{d'}{2} \right)^2 = \pi \left( \frac{3d}{2} \right)^2 = 9 \times \left( \pi \left( \frac{d}{2} \right)^2 \right) \implies A' = 9A$

Thus, the new resistance $R'$ is:

$R' = \frac{\rho L}{A'} = \frac{\rho L}{9A} = \frac{R}{9} \implies R' = \frac{1}{9} R$

The resistance decreases by a factor of 9.

RMS (root mean square) refers to the average value of an alternating current (a.c.) or voltage over a cycle. It provides a value that is equivalent to a direct current (d.c.) voltage that delivers the same power to a load.

A.C. (alternating current) is an electrical current that periodically reverses direction. Unlike d.c. (direct current), which flows constantly in a single direction, a.c. changes direction, delivering voltage that oscillates.

To find the total effective resistance ($R_{T}$) of the circuit, apply the formula for resistors in parallel:

$\frac{1}{R_{P}} = \frac{1}{R_{1}} + \frac{1}{R_{2}}$

Substituting the resistor values: $\frac{1}{R_{P}} = \frac{1}{200} + \frac{1}{500}$

Calculating:

$R_{P} = 143 \Omega$

Using Ohm's law, the current ($I_{T}$) flowing in the 200 Ω resistor can be calculated with the total voltage ($V$) and the effective resistance ($R_{T}$):

$I_{T} = \frac{V}{R_{T}} = \frac{12}{193} \approx 0.062 \text{ A}$

Removing the variable resistor will result in an increase in current through the circuit because the overall resistance is reduced. Thus the current flowing in the 50 Ω resistor will increase since it's now in a lower total resistance path.

The current that flows would decrease compared to the original circuit with the fixed 50 Ω resistor due to the characteristics of the coil which might introduce inductance.

The current flows in one direction only through the diode, which allows current in one way but blocks it in the reverse direction, thus preventing any current flow back.