Question

In the circuit shown in figure, E is a battery of emf $6V$ and internal resistance $1\Omega$. Find the reading of the ammeter A, if it has negligible resistance.

Answer 1

In the given circuit, $R_2,R_6$ and $R_4$ are in series. So,
$R_1^{\prime }=7+5+12=24\Omega$
Now $R_1^{\prime }$ and $R_5^{\prime }$ are in parallel. So,
$\frac{1}{R_2^{\prime }}=\frac{1}{8}+\frac{1}{24}=\frac{3+1}{24}=\frac{4}{24}=\frac{1}{6}$
$R_2^{\prime }=6$ohm.
Now $R_2^{\prime },R_1$ and $R_3$ are in series. So,
$R=R_2^{\prime }+R_1+R_3$
$=6+3+2=11$ohm.
We know $i=\frac{E}{R+r}=\frac{6}{11+1}=\frac{6}{12}=\frac{1}{2}$
$i=0.5$amp.