Question

Evaluate the current through the 1 2 resistor in the network shown in Fig. using mesh analysis 192 W 203 $32 1A $20 $ 10 V​

Answer 1

Answer:

I hope it's helpfull

Explanation:

Due to balanced wheatstone bridge ABCEA BE=8Ω can be taken out.

R AC

= (1+2)+(2+4),(1+2)(2+4) = 9

18=2

Now, ACFGHA is a balanced Wheatstone bridge, hence HC=10Ω can be taken out.

R AG

= 6+18,6∗18

= 24108

=4.5ΩR total

=2+4.5=6.5

Hence, current i= R total

V=6.5

6.5=1A