Question 75: Current Electricity

Question

The equivalent resistance between the points $A$ and $B$ in the following circuit is $\frac{x}{5} \Omega$. The value of $x$ is $\_\_\_\_$ .

Accepted Answer

The given circuit is a symmetric unbalanced bridge.
To find the equivalent resistance ( $R_{A B}$ ), we apply a potential difference $V$ across terminals $A$ and $B$ and determine the total current I entering the circuit.
Let the potential at node A be V (input terminal) and the potential at node B be 0.
Using Kirchhoff's Current Law (KCL) the sum of currents leaving each junction is zero.
At node $\mathrm{V}_1$ :
$$ \frac{\left(V_1-Vight)}{6}+\frac{\left(V_1-0ight)}{3}+\frac{\left(V_1-V_2ight)}{3}=0 $$
$$\Rightarrow $$ $\left(V_1-Vight)+2 V_1+2\left(V_1-V_2ight)=0$
$$\Rightarrow $$ $$ 5 V_1-2 V_2=V \ldots 	ext { (i) } $$
At node $\mathrm{V}_2$ :
$\frac{V_2-V}{3}+\frac{V_2-0}{6}+\frac{V_2-V_1}{3}=0$
$$\Rightarrow $$ $2\left(V_2-Vight)+V_2+2\left(V_2-V_1ight)=0$
$$\Rightarrow $$ $$ -2 \mathrm{~V}_1+5 \mathrm{~V}_2=2 \mathrm{~V} \ldots 	ext { (ii) } $$
Solving both the equations, (i) $	imes 5+$ (ii) $	imes 2$
$$ \left(5 V_1-2 V_2ight) 	imes 5+\left(-2 V_1+5 V_2ight) 	imes 2=V 	imes 5+2 V 	imes 2 $$
$$\Rightarrow $$ $\left(25 V_1-10 V_2ight)+\left(-4 V_1+10 V_2ight)=5 V+4 V$
$$\Rightarrow $$ $21 V_1=9 V \Rightarrow V_1=\frac{9}{21} V=\frac{3}{7} V$
Substituting $V_1$ into Equation (i) to find $V_2$ :
$$ \begin{gathered} 5\left(\frac{3}{7} \mathrm{~V}ight)-2 \mathrm{~V}_2=\mathrm{V} \ 2 \mathrm{~V}_2=\frac{15}{7} \mathrm{~V}-\mathrm{V}=\frac{8}{7} \mathrm{~V} \Rightarrow \mathrm{~V}_2=\frac{4}{7} \mathrm{~V} \end{gathered} $$
The total current I entering from terminal A is the sum of currents in the two left branches :
$$ I=\frac{V-V_1}{6}+\frac{V-V_2}{3} $$
$$\Rightarrow $$ $I=\frac{V-\frac{3}{7} V}{6}+\frac{V-\frac{4}{7} V}{3}$
$$\Rightarrow $$ $I=\frac{\frac{4 V}{7}}{6}+\frac{\frac{3 V}{7}}{3}=\frac{2 V}{21}+\frac{V}{7}$
$$ I=\frac{2 V+3 V}{21}=\frac{5 V}{21} $$
The equivalent resistance $R_{A B}$ is :
$$ R_{A B}=\frac{V}{I}=\frac{V}{5 V / 21}=\frac{21}{5} \Omega $$
Given that $\mathrm{R}_{\mathrm{AB}}=\frac{\mathrm{x}}{5} \Omega :$
$$ \frac{x}{5}=\frac{21}{5} \Rightarrow x=21 $$
Therefore, the value of x is 21.

Current Electricity

The equivalent resistance between the points A and B in the following circuit is \frac{x}{5} \Omega. The value of x is \_\_\_\_ .

Integer / Numeric Answer Type