Question

In the circuit shown in the figure, the ammeter reading is:

• A. 0.8 A
• B. 1.2 A
• C. 2 A
• D. 0.4 A

Answer 1

The correct option is A 0.8 A

$\text{Step 1: Simplify the given circuit by combining Series and Parallel Resistances}$

$\Rightarrow$

$\Rightarrow$

From figures $2$ and $3$, we see that both sets of $12\Omega$ and $4\Omega$ on either sides of ammeter are in parallel. So their equivalent resistances:

$\frac{1}{R_1}=\frac{1}{12}+\frac{1}{4}$ $=\frac{4}{12}$

$\Rightarrow R_1=3\Omega$

$\frac{1}{R_2}=\frac{1}{12}+\frac{1}{4}$ $=\frac{4}{12}$

$\Rightarrow R_2=3\Omega$

Now $R_1$ and $R_2$ are in series

So, $R^{\prime }=R_1+R_2=3+3=6\Omega$ $....\left(1\right)$

$\Rightarrow$

From Figures $4$ and $5$, $R^{\prime }$ and $4\Omega$ resistances are in parallel, So

$\frac{1}{R^{\prime \prime }}=\frac{1}{6}+\frac{1}{4}$ $=\frac{5}{12}$

$\Rightarrow R^{\prime \prime }=2.4\Omega$

Now, Equivalent resistance of the circuit

$R_{eq}=R^{\prime \prime }+0.6=2.4\Omega +0.6\Omega$

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

The Equivalent circuit is as follows:

$\text{Step 2: Apply Kirchhoff's Laws}$

Applying K.V.L in figure 6

Total current flowing from battery-

$I=\frac{V}{R_{eq}}$

$\Rightarrow I=\frac{6}{3}=2A$ $....\left(2\right)$

Applying KCL at rightmost junction in Figure $4$

$I=I_1+I_2$

$\text{Step 3: Solving Equations to find the required value}$

From figure $4$

Current in Ammeter(i.e. in Middle branch)

$I_1=I\frac{4\Omega }{4\Omega +R^{\prime }}=2\frac{4}{4+6}=0.8A$ (From equation $1$ and $2$)

Hence, $0.8A$ of current will flow through the Ammeter.