# 2 capicitor wire s habing diameter 3.0mm and T having diameter 1.5mm are connected in parallel a potential difference is applied across the ends of parallel arrangement the value of ratio of current T to current S is?

Jun 27, 2018

If I interpreted the question as intended, ${I}_{S} / {I}_{T} = \frac{\frac{V}{R} _ S}{\frac{V}{R} _ T} = {R}_{T} / {R}_{S} = 4$

#### Explanation:

I am not clear on first part of the description. I will assume:
2 wires of equal length, S having diameter 3.0mm and T having diameter 1.5mm are connected in parallel ...

Resistance of wire is inversely proportional to cross section area.

(The full formula for resistance of a length of wire is R = rho*l/A).

The properties $\rho \mathmr{and} l$ in both the wires are equal. Therefore the ratio of their resistances ${R}_{S} / {R}_{T}$ will simplify down to

${R}_{S} / {R}_{T} = \frac{\frac{1}{1.5 m m} ^ 2}{\frac{1}{0.75 m m} ^ 2} = \frac{1}{2} ^ 2 = \frac{1}{4}$

Considering Ohm's Law, the ratio of the currents

${I}_{S} / {I}_{T} = \frac{\frac{V}{R} _ S}{\frac{V}{R} _ T} = {R}_{T} / {R}_{S} = 4$

I hope this helps,
Steve