That depends on the damage to the meter.
Many current meters use a shunt resistor and measure the voltage drop across the shunt.
Initially, the leads to the movement may burn open or the movement winding may become shorted.
There is a very large ratio between the shunt impedance and the meter impedance.
To measure the voltage drop, the meter movement or digital circuit are connected in parallel with the shunt and share a small portion of the load current. Typically, a milli-voltmeter is used to measure the voltage drop.
If the movement or other components become shorted, the current share will increase until something burns open.
But, due to the very high ration between the shunt and the meter movement, there will be very little difference to the impedance that the shunt presents to the circuit being measured.
The impedance of the shunt is very low and it is unlikely that the shunt will short to a lower impedance.
In the event that the shunt fails open, the voltage across the open will rise to extreme levels.
Google "open CT secondary".
If the shunt fails open, there is a high probability of internal arcing and fire.
Bottom line:
If the shunt survives the overcurrent when the milli-voltmeter fails open, then there will be very little difference in the impedance presented to the circuit being metered.
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Ohm's law
Not just a good idea;
It's the LAW!
