Question 196: Current Electricity

Question

Two known resistances of $R\ \Omega$ and $2R\ \Omega$ and one unknown resistance $X\ \Omega$ are connected in a circuit as shown in the figure. If the equivalent resistance between points $A$ and $B$ in the circuit is $X\ \Omega$, then the value of $X$ is __________ $\Omega$.

Accepted Answer

Resistors 2R and $X$ are connected in series.
$$ \mathrm{R}_1=2 \mathrm{R}+\mathrm{X} $$
And the combination of $(2 R+X)$ is connected across $R$ in parallel.
$$ \frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{1}{\mathrm{R}_1}+\frac{1}{\mathrm{R}}=\frac{1}{2 \mathrm{R}+\mathrm{X}}+\frac{1}{\mathrm{R}} $$
$$\Rightarrow $$ $\frac{1}{X}=\frac{R+(2 R+X)}{R(2 R+X)}$
$$\Rightarrow $$ $X(3 R+X)=R(2 R+X)$
$$\Rightarrow $$ $3 R X+X^2=2 R^2+R X$
$$\Rightarrow $$ $$ X^2+2 R X-2 R^2=0 $$
Using the quadratic equation formula,
$$ X=\frac{-2 R \pm \sqrt{(2 R)^2-4(1)\left(-2 R^2ight)}}{2 	imes 1} $$
$$\Rightarrow $$ $X=\frac{-2 R \pm \sqrt{4 R^2+8 R^2}}{2}$
$$\Rightarrow $$ $X=\frac{-2 R \pm \sqrt{12 R^2}}{2}$
$$\Rightarrow $$ $X=\frac{-2 R \pm 2 \sqrt{3} R}{2}$
$$\Rightarrow $$ $X=-R \pm \sqrt{3} R$
$$\Rightarrow $$ $$ \mathrm{X}=(-1 \pm \sqrt{3}) \mathrm{R} $$
Since resistance cannot be negative, we take the positive root:
$$ \mathrm{X}=(\sqrt{3}-1) \mathrm{R} $$
Hence, the correct option is (C).

Answer

$2(\sqrt{3} - 1)R$

Answer

$R$

Answer

$(\sqrt{3} - 1)R$

Answer

$(\sqrt{3} + 1)R$

Current Electricity

Two known resistances of R\ \Omega and 2R\ \Omega and one unknown resistance X\ \Omega are connected in a circuit as shown in the figure. If the equivalent resistance between points A and B in the circuit is X\ \Omega, then the value of X is __________ \Omega.

Choose the correct answer: