Question

How do you solve this problem?

Answer 1

`V_2 = 24V, \\V_3 = 24V \\I _2 =4A\\I_3 = 2A\\I_1 = 6A\\R_1 = 1\Omega\\R_t = 5\Omega`

Step-by-step explanation:

The principle of continuity of current demands that

`I_1 = I_2+I_3`.

And applying Kirchhoff's current law two two loops in the circuit, we get:

(1).
`30V -I_1R_1 - I_2R_2 = 0`

and

(2).
`I_3R_3 -I_2R_2 = 0`.

Since
`I_1R_1 = 6V`, equation (1) becomes

`30V -6V - I_2R_2 = 0`

`\boxed{ 24V = I_2R_2}`

Since
`R_2 = 6\Omega`

`I_2 = (24V)/(6\Omega )`

`\boxed{I_2 = 4A}`

From equation (2) we now get:

`I_3R_3 = I_2R_2`

`I_3 = (I_2R_2)/(R_3)`

`I_3 = (6 A* 4\Omega)/(12\Omega)`

`\boxed{ I_3 = 2A.}`

Finally, we solve for
`I_1`

`I _1 = I_2+I_3`

`I_1 = 4A+2A`

`\boxed{ I_1 = 6A}`

therefore, the resistance
`R_1` is

`R_1 = (6V)/(6A) \\\\\boxed{ R_1 = 1\Omega}`

The potential drop
`V_2` is

`V_2 = I_2 R_2 \\\\V_2 = 4A*6\Omega\\\\\boxed{ V_2 = 24V. }`

Similarly, the potential drop
`V_3` is

`V_3 = I_3R_3 \\\\V_3 = 2A* 12\Omega\\\\\boxed{ V_3 = 24V}`

The total resistance of the circuit is
`R_t`

`R_t = R_1 +R_p`

where

`$(1)/(R_p) = (1)/(R_2)+(1)/(R_3) $`

`R_p = 4\Omega`;

therefore,

`R_t = 1\Omega+4\Omega`

`\boxed{ r_t =5\Omega.}`