Q8
Dear student
when AB is connected through wire 6 and 12 ohm resistors are in parallel to both sides hence net of the circuit will be 4+4 ohm=8
now using ohms law V=IR
80=I x 8
I = 10 A
now since voltage is constant across 12 and 6 ohm hence current I will divide into inverse ratio of resistance
I1 = 12/18 x 10 =20/3 A
I2 = 6/18 x 10 = 10/3 A
now net current flowing through wire AB = I1-I2 =20/3 -10/3 (downwards) =10/3 A.
Regards
when AB is connected through wire 6 and 12 ohm resistors are in parallel to both sides hence net of the circuit will be 4+4 ohm=8
now using ohms law V=IR
80=I x 8
I = 10 A
now since voltage is constant across 12 and 6 ohm hence current I will divide into inverse ratio of resistance
I1 = 12/18 x 10 =20/3 A
I2 = 6/18 x 10 = 10/3 A
now net current flowing through wire AB = I1-I2 =20/3 -10/3 (downwards) =10/3 A.
Regards