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# Current through the battery at the instance when the switch $S$ is closed in the given diagram is:

Last updated date: 22nd Jul 2024
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Hint:In electrical circuits, when capacitor and resistors are connected then we always check which two points of the circuit will get short circuited due to presence of capacitor and then will find total resistance of circuit and hence using ohm’s law of electricity, we will find the current through the circuit.

In the given circuit two capacitors of capacitance $2\mu F$ are connected and we can see the topmost capacitor will start to charge as soon as the circuit is closed when switch will be closed and hence the terminal between capacitor and resistance will get short circuited hence, no current will pass through them.

The circuit will have now only the battery of voltage $10V$ and a resistor of resistance $2\Omega$ . We know that, Ohm’s law of electricity said that resistance across the voltage is defined as $R = \dfrac{V}{I}$
Here, we have
$\text{Voltage} = 10\,V$
$R = 2\Omega$
Using the equation $R = \dfrac{V}{I}$
We get,
$I = \dfrac{{10}}{2}$
$\therefore I = 5\,A$
So, the current through the circuit is $5A$.

Hence, as soon as the switch $S$ is closed the current through the circuit will flow of magnitude $5A$.

Note: It’s important to remember that as the switch is closed and current starts to flow in the circuit the capacitor gets charged but as its direct current so the current will not pass through the capacitors is there were an alternating current, then there was a current across the capacitors which varies with time.