魅影直播

In the case of an idealized ammeter, there is zero resistance causing no potential

drop across it. This acts as a short circuit line, thereby disconnecting the$4.00\mathrm{惟}$

resistor from the circuit, and therefore the only resistance in the circuit now is the 2.00 鈩�

internal resistance of the battery.

According to Ohm鈥檚 law,

$$\begin{gathered} 蔚=Ir \\ I=\frac{蔚}{r} \end{gathered}$$

Substitute all the values in the above equation,

$$\begin{gathered} I=\frac{10.0\mathrm{V}}{2.00\mathrm{惟}} \\ =5.00\mathrm{A} \end{gathered}$$

Thus, the current is $5.00\mathrm{A}$.

The connection is parallel between the zero-resistance ammeter and also the 4.00 鈩� resistor. So, naturally all the current passes through the ammeter and therefore, there is no voltage drop across the 4.00 鈩� resistor.

Since there is no potential loss across the ammeter, therefore its terminal voltage is also zero.