Question

With reference to the diagram alongside, Calculate:

Reading of the ammeter.

Answer 1

Step 1- Given Data

First Resistance$R_1=4\Omega$.

Second Resistance$R_2=4\Omega$.

The potential difference of each cell$V=2V$.

Step 2 - To find out the equivalent resistance

As the resistors are connected in parallel-

The formula for equivalent resistance, $\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}$

$$\begin{gathered} \Rightarrow \frac{1}{R}=\frac{1}{4}+\frac{1}{6} \\ \Rightarrow \frac{1}{R}=\frac{5}{12} \\ \Rightarrow R=\frac{12}{5}=2.4\Omega . \end{gathered}$$

Thus, the equivalent resistance between P and Q is $2.4\Omega .$

Step 3 - To find out the current in the circuit

According to ohm's law, $V=I.R$, where ‘I’ is the current in the circuit.

The equivalent potential difference, $V=2+2=4V.$

$I=\frac{V}{R}$

$\Rightarrow I=\frac{4}{2.4}A=1.67A.$

Thus, the reading of the ammeter is$1.67A.$