Wire's Resistance

The electric field between the plates of a parallel-plate capacitor of capacitance 2.0 2.0 μ F \mu F drops to 1 3 \frac{1}{3} of its initial value in 4.4 4.4 μ s \mu s when the plates are connected by a thin wire.

Find the resistance of the wire.


The answer is 2.

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1 solution

The discharge characteristic of a capacitor as formulated below has been produced by considering a series R C RC circuit in which a capacitor at full charge is made to discharge at time, t = 0 t = 0 when the plates are connected through a resistor R R . We need to find this R R .

E = E 0 e t / R C E = E_{0} e^{-t/RC} ... (i)

E = 1 3 E 0 E = \frac{1}{3E_{0}} , C = 2.0 μ F C = 2.0 \mu F and t = 4.4 μ s t = 4.4 \mu s are given. Using these in (i), we get:

1 3 E 0 = E 0 e 4.4 / 2 R \frac{1}{3E_{0}} = E_{0} e^{-4.4/2R}

\implies l n ( 1 3 ) = e 4.4 / 2 R ln(\frac{1}{3}) = e^{-4.4/2R}

\implies 1.1 = 4.4 2 R -1.1 = -\frac{4.4}{2R}

\implies R = 2 Ω R = 2 \Omega (rounded off)

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