A pure resistance is connected as shown in the figure. The phase difference between the voltage applied and the current flowing in it will be

z e r o

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\frac{π}{2}

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- \frac{π}{2}

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\frac{π}{4}

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Solution

The correct option is A $z e r o$
Impedance of circuit $= R ∠ 0$

current $= \frac{V ∠ θ}{R ∠ 0} = \frac{V}{R} ∠ θ$

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Similar questions

List-I	List- II
(a) Phase difference between current and voltage in a purely resistive AC circuit	(i) π/2; current leads voltage
(b) Phase difference between current and voltage in a pure inductive AC circuit	(ii) zero
(c) Phase difference between current and voltage in a pure capacitive AC circuit	(iii) π/2; current lags voltage
(d) Phase difference between current and voltage in an LCR series circuit	(iv) ${tan}^{- 1} (\frac{X_{C} - X_{L}}{R})$

(\frac{1}{π})

Ω

\frac{π}{3}

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