$A$ coil,a capacitor,and an $AC$ source of $rms$ voltage $24 \,V$ are connected in series. By varying the frequency of the source,a maximum $rms$ current of $6 \,A$ is observed. If the coil is connected to a battery of $emf$ $12 \,V$ and internal resistance $4 \,\Omega$,then the current through it in the steady state is......$A$
- A$2.4$
- B$1.8$
- C$1.5$
- D$1.2$
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At a particular angular frequency,the reactance of a capacitor and that of an inductor are the same. If the angular frequency is doubled,the ratio of the reactance of the capacitor to that of the inductor will be:
DifficultMHT CET 2023
View SolutionIn an $A.C.$ circuit,the instantaneous $e.m.f.$ and current are given by:
$e = 100 \sin(20t)$
$i = 20 \sin(30t - \frac{\pi}{4})$
In one cycle of $A.C.$,the average power consumed by the circuit and the wattless current are,respectively:
$e = 100 \sin(20t)$
$i = 20 \sin(30t - \frac{\pi}{4})$
In one cycle of $A.C.$,the average power consumed by the circuit and the wattless current are,respectively:
The output of a step-down transformer is measured to be $48 \,V$ when connected to a $12 \,W$ bulb. The value of peak current is
Match List-$I$ with List-$II$ and choose the correct answer from the options given below:
List-$I$ List-$II$ $A$. Purely capacitive circuit $I$. $I$ leads $V$ by $90^{\circ}$ $B$. Purely inductive circuit $II$. $I$ and $V$ are in phase $C$. $LCR$ series at resonance $III$. $V$ leads $I$ by angle $\theta$ $D$. $LCR$ series circuit $IV$. $V$ leads $I$ by $90^{\circ}$
| List-$I$ | List-$II$ |
| $A$. Purely capacitive circuit | $I$. $I$ leads $V$ by $90^{\circ}$ |
| $B$. Purely inductive circuit | $II$. $I$ and $V$ are in phase |
| $C$. $LCR$ series at resonance | $III$. $V$ leads $I$ by angle $\theta$ |
| $D$. $LCR$ series circuit | $IV$. $V$ leads $I$ by $90^{\circ}$ |
DifficultJEE MAIN 2024
View SolutionAn $AC$ voltage source of $E = 150 \sin(100t)$ is used to run a device which offers a resistance of $20 \,\Omega$ and restricts the flow of current in one direction only. The $r.m.s.$ value of current in the circuit will be.....$A$
Difficult
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