[SOLVED] how much will be the current flow through the switch?

Hello, with respect to the below image how much will be current flow through the switch and resistor R2 when it is pressed?

assuming that the capacitor is fully charged and resistance offered by the switch is zero.

will it be 5V/10 = 0.5A. if that is the case then maximum current carried by the switch should be considered or not?

thanks

Current through switch=V1/(R1+R2), of course, after steady state is reached.

The simulator your using should tell you but I come up with 4.95 ma. Total resistance is
1000 ohms + 10 ohms = 1010 ohms.

Hi

and peak current will be 500mA. You correctly calculated it already.

Klaus

and peak current will be 500mA. You correctly calculated it already.

Click to expand...

No. The current will slowly rise from zero to the peak value already calculated in post #3, which is also the steady state value.

Hi,

No. The current will slowly rise from zero to the peak value already calculated in post #3, which is also the steady state value.

Click to expand...

Better do a simulation.

Klaus

Better do a simulation.

Click to expand...

Yes, yes, you are correct :thumbsup:

I don't see it. Please enlighten.

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At least spice sim doesn't see it.

Hi,

when closing the switch..
the current starts with 500mA
Then with a tau of about 10us the current reduces until it comes (close) to 4.95mA.

Klaus

Kajunbee said:

I don't see it. Please enlighten.

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At least spice sim doesn't see it.

Click to expand...

Let's analyze it together.

The capacitor voltage Vc will be 5V initially (assuming that the switch was left open long enough for C1 to charge fully like you said) and no current will be flowing in the circuit.

When the switch is closed, Vc will reduce gradually as C1 discharges till it settles at V1*R2/(R1+R2) = 49.5mV.

Realize that R2 is in parallel with C1, VR2 will always equal Vc.

In a nutshell:

Vc(initial) = V1 = 5V
Ir2(initial) = Vc(initial)/R2 = 500mA
Vc(final) = V1*R2/(R1+R2) = 49.5mV
Ir2 (final) = Vc(final)/R2 = 4.95mA

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KlausST said:

Hi,

when closing the switch..
the current starts with 500mA
Then with a tau of about 10us the current reduces until it comes (close) to 4.95mA.

Klaus

Click to expand...

Tau will be greater than R2*C1 = 10us because during Vc transient phase, V1 will also contribute current through R2 alongside that contributed by Vc but the voltage drop across R2 will maintain Vc. This means that C1 will discharge slower than it would without V1.

Also, owing to the fact that the current through R1 (let's call it Ir1) will start from 0A at Vc equals V1 and increase EXPONENTIALLY to (V1-Vc(final))/R1 = (5-0.0495)/1K = 4.95/1K = 4.95mA shows that tau will be much greater than 10us.

Hi,

almost perfect.

Tau will be greater than R2*C1 = 10us

Click to expand...

I´d say it tau will be smaller than R2 x C1, because the Rt that determines tau is the parallel connection of R1 and R2.

tau = Rt x C2
Where Rt = 1 / ( 1/R1 + 1/ R2 )
Rt will be smaller than R2.

****

This means that C1 will discharge slower than it would without V1.

Click to expand...

This is correct when you only look at dV/dt. But regarding tau this decreased dV/dt will be over-compensated by the now smaller voltage step to the target voltage of 49.5mV (against 0.0v).

Klaus