I have an arrangement in which an NMOS is connected in the ground path to control the power flow as shown below :-

enter image description here

When the FET is on , current will flow & there has to be voltage drop across the mosfet. Because it is connected on 0V already, how to analyze the voltage drop across it?

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    \$\begingroup\$ Look up its Rdson value and treat it as a resistor. \$\endgroup\$ – pjc50 Aug 5 '16 at 16:37

an NMOS is connected in the ground path the control the power flow

It's worth mentioning that when the FET is "off" current will flow through the diode - is this what you really want?

If so then just do as pjc50 says and treat the MOSFET as a resistor equal to its on-resistance.

Looking at your circuit a bit more, all it realistically does is protect the load from reverse polarity inputs.

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  • \$\begingroup\$ yes Andy, you are correct. My intent is to understand the voltage drop. If , i am driving with 5V/2A & ground is obviously @ 0V. What will be the voltage before & after the Mosfet if RDSon is 10mohm & MOSFET is in on state. \$\endgroup\$ – Oshi Aug 5 '16 at 16:47
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    \$\begingroup\$ @Rishi: Andy is telling you that your circuit will never turn off. Look at the diode in the FET. It will conduct when the FET is off. You have the FET in backwards. \$\endgroup\$ – Transistor Aug 5 '16 at 17:01
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    \$\begingroup\$ The FET isn't backwards, it's being used for polarity protection rather than switching. \$\endgroup\$ – Ignacio Vazquez-Abrams Aug 5 '16 at 17:03
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    \$\begingroup\$ @Rishi the loads most negative wire will be 20 mV higher than the incoming supply's most negative wire. \$\endgroup\$ – Andy aka Aug 5 '16 at 17:47
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    \$\begingroup\$ It's ohms law - just replace the MOSFET with a 10 milli ohm resistor in parallel with the diode and there you have it. \$\endgroup\$ – Andy aka Aug 5 '16 at 17:55

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