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(B) Given: $e = 75 \sin(\omega t) 	ext{ V}$ and $i = 1.5 \sin(\omega t + 45^\circ) 	ext{ A}$.Here,the current $i$ leads the voltage $e$ by a phase a

Vedclass · Accepted Answer

There must be an inductor in the box.

Vedclass · Answer

(B) Given: $e = 75 \sin(\omega t) 	ext{ V}$ and $i = 1.5 \sin(\omega t + 45^\circ) 	ext{ A}$.Here,the current $i$ leads the voltage $e$ by a phase angle of $\phi = 45^\circ$.Since the current leads the voltage,the circuit must be capacitive in nature. This implies that there must be a capacitor in the box.Since there is a phase difference between voltage and current,there must be a resistance $R$ in the box to provide the phase shift.Because the circuit is capacitive,the net reactance $X = X_C - X_L$ must be capacitive $(X_C > X_L)$. This does not strictly rule out the presence of an inductor,but it implies the net effect is capacitive. However,the statement 'There must be an inductor' is not necessarily true,making it the wrong statement.The power factor is $\cos \phi = \cos(45^\circ) = 1/\sqrt{2} \approx 0.707$. Thus,option $(d)$ is correct.Therefore,the wrong statement is $(b)$.

Similar Questions

In an $A.C.$ circuit,the current:

If $C, R, L$ and $I$ denote capacity, resistance, inductance and electric current respectively, the quantities having the same dimensions of time are :
$(1)$ $C R$
$(2)$ $\frac{L}{R}$
$(3)$ $\sqrt{L C}$
$(4)$ $L I^2$

In an $LCR$ series circuit,when $L$ is removed from the circuit,the phase difference between voltage and current is $\frac{\pi}{3}$. If $C$ is removed from the circuit instead of $L$,the phase difference is again $\frac{\pi}{3}$. The power factor of the circuit is $(\tan 60^{\circ}=\sqrt{3})$.

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