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

The resistance of a wire is inversely proportional to the cross-sectional area. When the diameter is increased by a factor of three, the radius is also increased by the same factor, leading to a change in the cross-sectional area (

$A = \pi r^2$

which becomes (A' = \pi (\frac{d}{2} \times 3)^2 = 9 \cdot A)). Hence, the new resistance will decrease. Mathematically:

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

This shows that the resistance decreases by a factor of nine.

RMS stands for root mean square and it represents the effective value of an alternating current (a.c.) waveform. It provides a way of expressing the average power of the current, effectively allowing it to be compared to direct current (d.c.). The rms voltage is the average of the squares of instantaneous values divided by time, resulting in an equivalent d.c. voltage that would produce the same power in a resistive load.

A.C. stands for alternating current, which is a type of electrical current that periodically reverses direction. Unlike direct current (d.c.), which flows in a single direction, a.c. changes its direction, usually in a sinusoidal waveform. This is the form of current most commonly used in power supplies and is characterized by its frequency (in Hertz) and amplitude.

To find the effective resistance of the circuit, we use the formula for resistors in parallel:

$\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2}$

Here, R1 = 200 Ω and R2 = 500 Ω. Thus:

$\frac{1}{R_p} = \frac{1}{200} + \frac{1}{500}$

Calculating it gives:

$R_p = 143.0 \text{ Ω}$

Therefore, the effective resistance of the circuit is approximately 143 Ω.

To find the current flowing in the 200 Ω resistor, we apply Ohm's Law:

$I = \frac{V}{R}$

where V = 12 V and R = 200 Ω. So:

$I_{200} = \frac{12}{200} = 0.06 \text{ A}$

Thus, the current in the 200 Ω resistor is 0.06 A.

Removing the variable resistor will reduce the total resistance of the circuit, allowing more current to flow. Since the total current in a parallel circuit is split among the resistors, the current through the 50 Ω resistor will increase as it receives a higher share of the total current due to the decreased total resistance of the circuit.

Replacing the 50 Ω resistor with a coil of the same resistance will have minimal impact on the current, as the resistance remains the same. However, the coil may introduce some inductance which can affect the current flow depending on the characteristics of the circuit.

Replacing the 50 Ω resistor with a diode will significantly affect the current flow. The diode allows current to flow in one direction only, effectively blocking the current flow in the reverse direction. This will usually lead to a reduction in the overall current flowing in the circuit as compared to a resistor that allows bidirectional current.