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

 

 
 
 
 
 

Answers (1)
S safeer

 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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