Question

caluclate the current 'I' in the circuit shown in the fig ​

Answer 1

Answer:

1.2A

Step-by-step explanation:

6and 24 are in series

3 and 12 are in series

R1=6+24=30 ohm

R2=3+12=15ohm

R1 and R2 are in parallel

1/R=1/30+1/15

1/R=3/10

R=10ohm

Now V=IR

I=V/R

I=12/10=1.2A

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