Question

How do you solve this problem?

Answer 1

Answer:

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

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 $R_2 = 6\Omega$

$I_2 = \dfrac{24V}{6\Omega }$

$\boxed{I_2 = 4A}$

From equation (2) we now get:

$I_3R_3 = I_2R_2$

$I_3 = \dfrac{I_2R_2}{R_3}$

$I_3 = \dfrac{6 A* 4\Omega}{12\Omega}$

$\boxed{ I_3 = 2A.}$

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

$R_1 = \dfrac{6V}{6A} \\\\\boxed{ R_1 = 1\Omega}$

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

$\frac{1}{R_p} = \frac{1}{R_2}+\frac{1}{R_3}$

$R_p = 4\Omega$;

therefore,

$R_t = 1\Omega+4\Omega$

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