
Figure shows a circuit that contains three identical resistors with resistance $R = 9.0\Omega $ each, two identical inductors with inductance $L = 2.0mH$ each, and an ideal battery with emf$\varepsilon = 18V$. The current $i$ through the battery just after the switch closed is:

$\left( a \right)$ $0.2A$
$\left( b \right)$ $2A$
$\left( c \right)$ $0A$
$\left( d \right)$ $2mA$
Answer
148.8k+ views
Hint As we know in the DC source capacitor acts as an open circuit and the inductor will act as a short circuit. So we have the equivalent circuit as a $18V$ source and only there is a resistance whose value is$9.0\Omega $. So from here, we can easily find them by using the current formula.
Formula used
Current will be equal to
$I = \dfrac{V}{R}$
Here,
$I$, will be the current
$V$, will be the voltage
$R$, will be the resistance
Complete Step By Step Solution
So in this question, we have an identical resistor and also there are two inductors. There is an ideal battery connected through it. We have to find the current through the battery when the switch is being closed.
When the switch is ON, that is at$t = 0$,
The capacitor will behave as a short circuit and also the inductor will behave like an open circuit.
Due to this, the inductor will be zero and only ${R_2}$will carry the resistance.
Therefore by using the current formula
Current will be equal to
$I = \dfrac{V}{R}$
From the question, it can be written as
$ \Rightarrow \dfrac{\varepsilon }{{{R_2}}}$
Now substitute the values, we get
$ \Rightarrow \dfrac{{18}}{9}$
On furthermore calculation, we get
$ \Rightarrow 2A$
Therefore, the current $2A$ is required. Hence the option $B$will be correct.
Note An inductor is an inactive electronic part that stores energy as an attractive field. An inductor comprises a wire circle or loop. It is as a rule as a loop, twisted over an attractive or even air at its center. Indeed, even a straight wire creates an attractive field and can be considered as an inductor. The inductance is legitimately corresponding to the number of turns in the curl.
Formula used
Current will be equal to
$I = \dfrac{V}{R}$
Here,
$I$, will be the current
$V$, will be the voltage
$R$, will be the resistance
Complete Step By Step Solution
So in this question, we have an identical resistor and also there are two inductors. There is an ideal battery connected through it. We have to find the current through the battery when the switch is being closed.
When the switch is ON, that is at$t = 0$,
The capacitor will behave as a short circuit and also the inductor will behave like an open circuit.
Due to this, the inductor will be zero and only ${R_2}$will carry the resistance.
Therefore by using the current formula
Current will be equal to
$I = \dfrac{V}{R}$
From the question, it can be written as
$ \Rightarrow \dfrac{\varepsilon }{{{R_2}}}$
Now substitute the values, we get
$ \Rightarrow \dfrac{{18}}{9}$
On furthermore calculation, we get
$ \Rightarrow 2A$
Therefore, the current $2A$ is required. Hence the option $B$will be correct.
Note An inductor is an inactive electronic part that stores energy as an attractive field. An inductor comprises a wire circle or loop. It is as a rule as a loop, twisted over an attractive or even air at its center. Indeed, even a straight wire creates an attractive field and can be considered as an inductor. The inductance is legitimately corresponding to the number of turns in the curl.
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