Question

A circuit with resistors, a cell and an ammeter is shown in the figure. Find the value of current coming out from the battery and also find the ammeter reading.

Answer 1

The circuit in the question can be simplified to figure 1.1, which can be further be simplified as figure 1.2.
From the figure 1.2, we can see, that the three $4\Omega$ resistors are in parallel and this parallel combination is in series with the other $4\Omega$ resistor and the battery.
The Equivalent resistance of the parallel combination is given by
$\frac{1}{R_{eq}}=\frac{1}{4}+\frac{1}{4}+\frac{1}{4}=\frac{3}{4}$

$\Rightarrow R_{eq}=\frac{4}{3}\Omega$

This simplifies the circuit as shown in figure 1.3
Now, the $R_{eq}$ and the $4\Omega$ resistor are in series. So, the equivalent resistance for this combination is given by,
$R_s=R_{eq}+4=\frac{4}{3}+4=\frac{16}{3}\Omega$

So the circuit simplifies to figure 1.4

Using Ohm's Law, we have

$V=iR$

$2=i\times \frac{16}{3}$

$\Rightarrow \frac{2}{\frac{16}{3}}=i$

$\Rightarrow i=\frac{3}{8}A$

Now, the current in the ammeter will be $\frac{i}{3}A$ because the current entering the parallel branch is $i$ and since all the branches are having equal resistances, the current divides equally in all the branches.

Thus, the ammeter reading will be $i_A=\frac{1}{3}\times \frac{3}{8}=\frac{1}{8}A$

Hence, the current in the circuit, $i$ is $\frac{3}{8}A$ and the ammeter reading is $\frac{1}{8}A$.