# Evan's Space

## Brightness of bulb as resistance of rheostat increases

Solution:A

As resistance of the rheostat increases, the total resistance in parallel will increase. Hence potential difference across the parallel bulb Q increases, therefore BRIGHTER.

Total effective resistance of the circuit (include resistance of bulb P) increases, main current thought circuit (bulb P) decreases. Hence bulb P DIMMER.

You can think in terms of p.d of bulb P. Since p.d. across parallel branches increases, p.d across bulb P decreases, hence bulb P is DIMMER.

Without calculation, using resistance and p.d. is easier. To convince yourself, refer to the following diagrams with imaginary numbers.

## N2007P1Q27 – Current Electricity – Brightness of bulbs

X and Y are lamps with filaments made of the same material. The filament of lamp X is thicker and shorter than that of lamp Y. Each lamp is connected to the mains and switched on. Which is the brighter lamp and which lamp has the larger resistance?

 Brighter lamp Larger resistance A X X B X Y C Y X D Y Y

Solutions: Option B

Using the formula, R = ρl/A     (ρ: resistivity (Ωm), l: length (m) and A: cross-sectional area (m2) Since both are of the same material, resistivity ρ is the same. X is thicker (cross-sectional area) and shorter, X will have lower resistance than Y.

Hence Y has larger resistance. Both are then connected to mains.

You can assume they are connected individually as shown below. Potential difference across each bulb is the same as the mains.

Since brightness depends on power, using P = IV, bulb X is brighter.