Question

i would like some help with this problem, i already tried to complete it on my own but would like to check my answer

Answer 1

Given:

Two circuits with multiple resistors

To find:

The net resistance of the circuits

Step-by-step explanation:

For the first circuit

The equivalent series resistance of the 1 and 2 is,

`\begin{gathered} R_1+R_2 \\ =20+16 \\ =36\text{ ohm} \end{gathered}`

The equivalent series resistance of the 4 and 5 is,

`\begin{gathered} 10+14 \\ =24\text{ ohm} \end{gathered}`

The net resistance of the resistances is,

`\begin{gathered} (1)/(R)=(1)/(36)+(1)/(24)+(1)/(12) \\ (1)/(R)=(11)/(72) \\ R=(72)/(11)\text{ohm} \\ R=6.54\text{ ohm} \end{gathered}`

Hence, the net resistance of the upper circuit is 6.54 ohms.

For the circuit below:

`R_2,\text{ R}_3,\text{ R}_4\text{ are in series}`

So, the equivalent resistance is,

`\begin{gathered} R_2+R_3+R_4 \\ =2+2+2 \\ =6\text{ ohm} \end{gathered}`

This equivalent resistance in parallel with

`R_5`

So, the equivalent resistance is,

`\begin{gathered} (6*2)/(6+2) \\ =1.5\text{ ohm} \end{gathered}`

Now 1.5 ohm is in series with the rest two resistances.

So, the net resistance is,

`\begin{gathered} R_1+R_6+1.5 \\ =2+2+1.5 \\ =5.5\text{ ohm} \end{gathered}`

Hence, the net resistance of the circuit below is 5.5 ohms.