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Resistances R1, R2, R3, R4, R5, and R6 a...

Resistances `R_1, R_2, R_3, R_4, R_5, and R_6` are connected with a 6V battery with internal resistacne `1Omega` as shown in fig 5.90. Find (a) equivalent resistance and (b) current in each resistacne.

Text Solution

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Hence, the equivalent resistance between
A and B is
`(1)/(R_(AB)) = (1)/(16) + (1)/(8) +(1)/(16) = (4)/(16)`
or `R_(AB) = 4Omega`
Now the circuit reduces to the one shown in. The
current supplied by the battery is
`I = (60)/(4 +1) = 12A`
As potential difference across `R_5` is zero, no current wil
be in `R_5`. Now using current distribution law, we can find the
current in each branch.
the currentsw in `R_1 and R_2` will be equal.
`I_1 = I((R'_2R'_3)/(R'_1R'_2 +R'_2R'_3 +R'_3R'_1))`
`=12[(8xx16)/(16xx8+8xx16+16xx16)] =3A`
The currents in `R_3 and R_4`is
`I_2 =I((R'_1R'_3)/(R'_1R'_2 + R_3'R_1'))`
`= 12 [(16 xx 16)/(16 xx 8 + 8 xx 16 + 16 xx 16)] = 6 A`
Hence, current in `R_6` is `I_3` `= 12 - (3+6) = 3A` .
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