Answer:
The current in the branch AB will be 0.66 Amperes
Explanation:
form the current division rule:
current in the 1st 3Ω resistors in branch PA
= 2×6/9
=4/3A
current in the 1st 6Ω resistor in branch AB
=2×3/9
=2/3A
Again these current will divide to the second parallel set of resistor
Hence current 2nd 3 Ω resistors in branch BQ from the branch PA
=4/3×6/9
=8/9A
similarly current 2nd 6Ω resistor in branch Aa from the branch PB
=2/3×3/9
=2/9A
Hence current in the branch AB
=8/9-2/9
=6/9
=0.66Aa
Hence the current in the branch AB will be 0.66 apmeres.
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Answers & Comments
Answer:
The current in the branch AB will be 0.66 Amperes
Explanation:
form the current division rule:
current in the 1st 3Ω resistors in branch PA
= 2×6/9
=4/3A
current in the 1st 6Ω resistor in branch AB
=2×3/9
=2/3A
Again these current will divide to the second parallel set of resistor
Hence current 2nd 3 Ω resistors in branch BQ from the branch PA
=4/3×6/9
=8/9A
similarly current 2nd 6Ω resistor in branch Aa from the branch PB
=2/3×3/9
=2/9A
Hence current in the branch AB
=8/9-2/9
=6/9
=0.66Aa
Hence the current in the branch AB will be 0.66 apmeres.