Question

Seven resistors are connected as shown in the diagram: The equivalent resistance in ohms of this network between $A$ and $B$ is:

A. $6\mathrm{\Omega }$
B. $8\mathrm{\Omega }$
C. $12\mathrm{\Omega }$
D. $20\mathrm{\Omega }$

Answer 1

Hint: : Identify the parallel and series connection in the given circuit. First solve for the resistors in parallel connection, then the circuit will become a Wheatstone bridge. Check the condition for a balanced Wheatstone bridge.

Formula used:
Resistors connected in series, equivalence resistance $R_{eq}=R_1+R_2+R_3+...$
Resistors connected in parallel, equivalence resistance ${\displaystyle \frac{1}{R_{eq}}}={\displaystyle \frac{1}{R_1}}+{\displaystyle \frac{1}{R_2}}+{\displaystyle \frac{1}{R_3}}+...$

Complete step by step answer:
First name the resistances and redraw the circuit diagram.

In the above circuit, we have to find the equivalent resistance between the terminal A and B.The resistors $R_6$ and $R_7$ are connected in parallel.Equivalence of $R_6$ and $R_7$ is given as
$R_{67}={\displaystyle \frac{R_6\cdot R_7}{R_6+R_7}}$
Substitute the values of $R_6$ and $R_7$ in the above formula.
$\Rightarrow R_{67}={\displaystyle \frac{8\times 10}{8+10}}\mathrm{\Omega }$
$\Rightarrow R_{67}={\displaystyle \frac{40}{9}}\mathrm{\Omega }$
The resistors $R_1$ and $R_5$ are connected in parallel.
The equivalent resistor of $R_1$ and $R_5$ is given as
$R_{15}={\displaystyle \frac{R_1\cdot R_5}{R_1+R_5}}$
Substitute the values of $R_1$ and $R_5$ in the above formula.
$\Rightarrow R_{15}={\displaystyle \frac{30\times 6}{30+6}}\mathrm{\Omega }$
$\Rightarrow R_{15}=5\mathrm{\Omega }$

Now the equivalent circuit of above the circuit is as follows

The equivalent circuit of the given circuit is a Wheatstone bridge circuit.
The Wheatstone bridge is balanced because it satisfies the condition ${\displaystyle \frac{R_{15}}{R_4}}={\displaystyle \frac{R_2}{R_3}}$
Therefore, no current flows through the resistor $R_{67}$. Remove the $R_{67}$ from the Wheatstone bridge. The final circuit is as follows.

$R_{15}$ and $R_2$ are in series.
The equivalent resistance of $R_{15}$ and $R_2$ is $R_{152}=R_{15}+R_2$
$\Rightarrow R_{152}=5\mathrm{\Omega }+4\mathrm{\Omega }=9\mathrm{\Omega }$
$R_4$ and $R_3$ are in series connection.
The equivalent resistance of $R_4$ and $R_3$ is $R_{45}=R_4+R_5$
$\Rightarrow R_{45}=10\mathrm{\Omega }+8\mathrm{\Omega }=18\mathrm{\Omega }$
Now $R_{152}$ and $R_{45}$ are in parallel connection.
The equivalent resistance of $R_{152}$ and $R_{45}$ is $$R_{eq}={\displaystyle \frac{R_{152}\cdot R_{45}}{R_{152}+R_{45}}}$$.
Substitute the values of $R_{152}$ and $R_{45}$ in the above formula.
$\Rightarrow R_{eq}={\displaystyle \frac{9\times 18}{9+18}}\mathrm{\Omega }$
$\therefore R_{eq}=6\mathrm{\Omega }$
The equivalent resistance of the given circuit is $6\mathrm{\Omega }$.

Hence the correct option is A.

Note: Alternative method:Alternatively, we can solve the circuit by drawing an equivalent circuit of the given circuit with some extra consideration as follows.

To solve the above equivalence circuit, Let the current $I$ enter junction A and leave junction B. A battery of $V$ voltage is connected across the terminals A and B. For simplification let terminal A is at $V$ voltage and terminal B is at $0$ voltage. Let the junctions C and D are at $V_1$ voltage and $V_2$ voltage respectively. Now apply Kirchhoff's current law at the junction A in the above equivalence circuit.
$I=I_1+I_2$

Apply Kirchhoff's current law at the junction C in the above equivalence circuit.
$I_1=I_3+I_4$
From the above circuit diagram, $I_1={\displaystyle \frac{V-V_1}{5}}$
$I_3={\displaystyle \frac{V_1-0}{4}}$
$\Rightarrow I_3={\displaystyle \frac{V_1}{4}}$
$\Rightarrow I_4={\displaystyle \frac{V_1-V_2}{{\displaystyle \frac{40}{9}}}}$
$\Rightarrow I_4={\displaystyle \frac{9}{40}}\left(V_1-V_2\right)$
Therefore,
$${\displaystyle \frac{V-V_1}{5}}={\displaystyle \frac{V_1}{4}}+{\displaystyle \frac{9}{40}}\left(V_1-V_2\right)$$
On simplification
$\Rightarrow 27V_1-9V_2=8V$ …… (1)
Now apply Kirchhoff's current law at junction D in the above equivalence circuit.
$I_2+I_4=I_5$

From the above circuit diagram, $I_2={\displaystyle \frac{V-V_2}{10}}$
$I_5={\displaystyle \frac{V_2-0}{8}}$
$\Rightarrow I_5={\displaystyle \frac{V_2}{8}}$
Therefore,
$${\displaystyle \frac{V-V_2}{10}}+{\displaystyle \frac{9}{40}}\left(V_1-V_2\right)={\displaystyle \frac{V_2}{8}}$$
On simplification
$\Rightarrow -9V_1+18V_2=4V$ …… (2)
Now solve the two equations (1) and (2) and find the values of $V_1$ and $V_2$ in terms of $V$. We got,
$V_1={\displaystyle \frac{4V}{9}}$,
$\Rightarrow V_2={\displaystyle \frac{4V}{9}}$

Now calculate the value of $I_1={\displaystyle \frac{V-V_1}{5}}$
$I_1={\displaystyle \frac{V-{\displaystyle \frac{4V}{9}}}{5}}$
$\Rightarrow I_1={\displaystyle \frac{V}{9}}$
And calculate the value of $I_2={\displaystyle \frac{V-V_2}{10}}$
$I_2={\displaystyle \frac{V-{\displaystyle \frac{4V}{9}}}{10}}$
$\Rightarrow I_2={\displaystyle \frac{V}{18}}$
Now we have $I=I_1+I_2$
Substitute the values of $I_1$ and $I_2$ in the above formula for $I$.
$I={\displaystyle \frac{V}{9}}+{\displaystyle \frac{V}{18}}$

Further calculating
$\Rightarrow I={\displaystyle \frac{V}{6}}$
The equivalent resistance of the above equivalence circuit between the two terminals A and B is given by
$R_{eq}={\displaystyle \frac{V_B-V_A}{I}}$
Substitute all the required values in the above formula
$R_{eq}={\displaystyle \frac{V-0}{{\displaystyle \frac{V}{6}}}}$
On further simplification
$R_{eq}=6\mathrm{\Omega }$
Hence the correct option is (A).