In the given below, find the voltmeter reading across a 4 \Omega  resistor.

 

 
 
 
 
 

Answers (1)
S safeer

In the second loop, three resistors are in series
R_{1}= 2+4+2= 8\Omega
Now above equivalent resistance is parallel to the 8-ohm resistor
\frac{1}{R_{2}}= \frac{1}{8}+\frac{1}{8}
\frac{1}{R_{2}}= \frac{2}{8}= \frac{1}{4}
R= 4\Omega
Now,
R_{2} in series with 2\Omega and2\Omega
Hence,
the equivalent resistance\left ( R_{equi} \right )= 2+4+2= 8\Omega

The current from the battery would be

=\frac{v}{R_{1}} 
\therefore \frac{16}{8}= 2Ampere.
The total current in the circuit is 2A. 

Now 1 A flows through 8 ohm and 1 A flows through the series resistors (4+2+2=8 ohm)

Now voltage across 4 ohm is

V=IR=1\times4=4V

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