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?

1 Answer
Steve J.
Jun 27, 2018

If I interpreted the question as intended, #I_S/I_T = (V/R_S)/(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 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 =(1/(1.5 mm)^2)/(1/(0.75 mm)^2) = 1/2^2 = 1/4#

Considering Ohm's Law, the ratio of the currents

#I_S/I_T = (V/R_S)/(V/R_T) = R_T/R_S = 4#

I hope this helps,
Steve

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