Question

An idealized ammeter is connected to a battery as shown in Fig.

E25.28. Find (a) the reading of the ammeter, (b) the current through the$\mathbf{4}\mathbf{.}\mathbf{00}\mathbf{\Omega }$

resistor, (c) the terminal voltage of the battery.

Fig. E25.28.

Short answer

a) The reading of the ammeter is 5.00 A.

b) The current through the $4.00\mathrm{\Omega }$resistor is zero.

c) The terminal voltage of the battery is zero.

Step-by-step solution

 (a) Determination of the reading of the ammeter.

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$4.00\mathrm{\Omega }$

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

internal resistance of the battery.

According to Ohm’s law,

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

Substitute all the values in the above equation,

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

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

(b) Determination of the current through the 4.00 Ω resistor.

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

(c) Determination of the terminal voltage of the battery.

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