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user1844
user1844

Presumably theThe two 'relations' are A. between the burden voltage of their ammeter and your load voltage, and B. between the leakage current of their voltmeter and your load current. It's not really the ratio of voltage to current in your load, although obviously if that was extreme then is would imply very low voltage or very low current.

My feeling is that at the currents/voltages you're operating at, it won't make much difference either way. If you were measuring some micropower single-coin-cell-for-three-years job, then you'd have to care.

Presumably the two 'relations' are A. between the burden voltage of their ammeter and your load voltage, and B. between the leakage current of their voltmeter and your load current.

My feeling is that at the currents/voltages you're operating at, it won't make much difference either way. If you were measuring some micropower single-coin-cell-for-three-years job, then you'd have to care.

The two 'relations' are A. between the burden voltage of their ammeter and your load voltage, and B. between the leakage current of their voltmeter and your load current. It's not really the ratio of voltage to current in your load, although obviously if that was extreme then is would imply very low voltage or very low current.

My feeling is that at the currents/voltages you're operating at, it won't make much difference either way. If you were measuring some micropower single-coin-cell-for-three-years job, then you'd have to care.

Source Link
user1844
user1844

Presumably the two 'relations' are A. between the burden voltage of their ammeter and your load voltage, and B. between the leakage current of their voltmeter and your load current.

My feeling is that at the currents/voltages you're operating at, it won't make much difference either way. If you were measuring some micropower single-coin-cell-for-three-years job, then you'd have to care.