Question

10. Derive an expression for equivalent resistance in the following case. Decide which resistanes are in series and parallel. Solve for series and then for parallel Combine both the results to get the equivalent resistance.​

Answer 1

R2 and R3 are in series. R4 and R5 are in series. (R2+R3) and (R4+R5) are in parallel.

The equivalent resistance is:

$R_{eq}=R_1+ \frac{(R_2+R_3)(R_4+R_5)}{R_1+R_2+R_3+R_4}$

Answer 2

Answer:

The resistances $R_{4}$ and $R_{5$ are in series,

The resistances $R_{2}$ and $R_{3}$ are in series,

The expression for equivalent resistance is $\frac{R_{45}R_{23}}{R_{23}+R_{23}} +R_{1}$.

Explanation:

Here,

The equivalent resistance of $R_{2}$ and $R_{3}$ is denoted by $R_{23}$.

The equivalent resistance of $R_{4}$ and $R_{5$ is denoted by $R_{45}$.

The equivalent resistance of $R_{23}$ and $R_{45}$ is denoted by $R_{2345}$.

The equivalent resistance of $R_{2345}$ and $R_{1}$ is denoted by $R_{eq}$.

Now,

The resistances $R_{2}$ and $R_{3}$ are in series,

By the equation,

$R_{23}$ =  $R_{2}$ + $R_{3}$

Then,

The resistances $R_{4}$ and $R_{5$ are in series,

By the equation,

$R_{45}$ =  $R_{4}$ + $R_{5$

Then,

The resistances $R_{23}$ and $R_{45}$ are in parallel,

By the equation,

$\frac{1}{R_{2345}} =\frac{1}{R_{23}} +\frac{1}{R_{45} } \\R_{2345}=\frac{R_{45}R_{23}}{R_{23}+R_{23}}$

Then,

The resistances $R_{2345}$ and $R_{1}$ are in series,

By the equation,

$R_{eq}= R_{2345}+R_{1} \\R_{eq}=\frac{R_{45}R_{23}}{R_{23}+R_{23}} +R_{1}$

So,

The resistances $R_{4}$ and $R_{5$ are in series,

The resistances $R_{2}$ and $R_{3}$ are in series,

The expression for equivalent resistance is $\frac{R_{45}R_{23}}{R_{23}+R_{23}} +R_{1}$.