Is there any relationship between the current and the cross-section of the wire? If the material is the same and the length is the same, only the cross-sectional area is different I = nesv

Is there any relationship between the current and the cross-section of the wire? If the material is the same and the length is the same, only the cross-sectional area is different I = nesv

In terms of current carrying capacity, the larger the cross-sectional area, the better. Because the cross-sectional area is small, the resistance will increase, and the loss power of this section of line will increase, P = I ^ 2 * r

Is there a relationship between wire length and current?
If the real number of ammeter is larger, the voltmeter will be smaller?

The length of the wire has nothing to do with the current. The current I = u / R, the size of R generally does not consider the resistance of the wire. If it is considered, the resistance of the wire is inversely proportional to the cross section and directly proportional to the length of the wire
By the way, under what circumstances, the larger the real number of ammeter, the smaller the voltmeter?
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In the case of a certain power consumption

As shown in the figure, AB and BC are two conductors made of the same material with the same length and different cross-sectional area. After they are connected in series, they are connected into the circuit. The current through them is compared______ (select "IAB < IBC", "IAB = IBC" or "IAB > IB")

When AB and BC are connected in series, the current through them is equal, that is, IAB = IBC. So the answer is: IAB = IBC

In a series circuit, if a light bulb is connected in series, how will the total voltage and current and the voltage and current of the original light bulb change?
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The total voltage and current remain unchanged
The voltage at both ends of the original bulb is reduced, because adding a bulb has a considerable amount of resistance, the resistance is large, the current in the circuit is reduced, but the resistance of the original bulb remains unchanged, u = IR I is small, the voltage is naturally less
The current is also small, as explained above