# In the circuit shown in the figure, find the value of the current shown in the ammeter A.

figure 1

to find the current shown in ammeter, find the current through 6$\Omega$ resistance. Here assume ammeter is ideal, so resistance of ammeter is is zero. The 4$\Omega$ resistance and 2$\Omega$ resisatnce are in series.

We know at series connection $R_{equivalent}= R_{1}+R_{2}$

parallel connection $R_{equivalent}=\frac{R_{1}R_{2}}{R_{1}+R_{2}}$

So $R= 2\Omega +4\Omega =6\Omega$

The figure can be redrawn into

figure 2

$6\Omega$ and $6\Omega$ are parallel so effective resistance is $\frac{6\times 6 }{6+6}=\frac{36}{12}= \frac{6}{2}= 3\Omega$

Then figure become

figure 3

9$\Omega$ and 3$\Omega$ are in series,

effective resistance is 9 $\Omega$ + 3 $\Omega$ = 12 $\Omega$

Figure become,

figure 4

12 $\Omega$ and 12 $\Omega$ are in parallel so,

$\frac{12\times 12 }{12+12}=\frac{12}{2}=6\Omega$

figure 5

3$\Omega$ and 6 $\Omega$ are in series, so

3$\Omega$ + 6$\Omega$ = 9 $\Omega$

By ohm's law,

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

$I = \frac{9V}{9\Omega }= 1A$

In figure 4 both 12 $\Omega$ resistance have $\frac{I}{2}$ current = 0.5A

The equivalent figure is,

from the above diagram, the current through 6$\Omega$ resistance is $\frac{I}{4}=0.25A$

Hence, the current shown in the ammeter is 0.25A,

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