Question

in parallel circuits, the total resistance of a circuit is determined by using the formula $\frac{1}{r_t}=\frac{1}{r_1}+\frac{1}{r_2}$, where $r_t$ is the total resistance, $r_1$ is the resistance of one branch of the parallel circuit, and $r_2$ is the resistance of the other branch of the parallel circuit. in a particular parallel circuit, one branch has $5omega$ more resistance than the other. this can be modelled by the equation $\frac{1}{r_t}=\frac{1}{x}+\frac{1}{x + 5}$ where $x$ is the resistance of one branch of the parallel circuit, in ohms. if the total resistance of this circuit, $r_t$, is $6omega$, then the values for the resistance of the two branches of the circuit are
$5omega$ and $10omega$
$10omega$ and $15omega$
$3omega$ and $8omega$
$3omega$ and $3omega$

Explanation:

Step1: Substitute $R_T = 6$ into the formula

Substitute $R_T=6$ into $\frac{1}{R_T}=\frac{1}{x}+\frac{1}{x + 5}$, we get $\frac{1}{6}=\frac{1}{x}+\frac{1}{x + 5}$.

Step2: Find a common - denominator

The common denominator of the right - hand side is $x(x + 5)$. So $\frac{1}{6}=\frac{x + 5+x}{x(x + 5)}$, which simplifies to $\frac{1}{6}=\frac{2x + 5}{x(x + 5)}$.

Step3: Cross - multiply

Cross - multiplying gives $x(x + 5)=6(2x + 5)$. Expand to get $x^{2}+5x=12x + 30$.

Step4: Rearrange to a quadratic equation

Rearrange to $x^{2}+5x-12x - 30 = 0$, so $x^{2}-7x - 30=0$.

Step5: Factor the quadratic equation

Factor $x^{2}-7x - 30$ as $(x - 10)(x+3)=0$.

Step6: Solve for $x$

Setting each factor equal to zero gives $x-10 = 0$ or $x + 3=0$. Since resistance cannot be negative, we take $x = 10$.

Step7: Find the two resistances

If $x = 10$, then the other resistance is $x + 5=15$.

Answer:

$10\Omega$ and $15\Omega$