Question

find the items on the circuit, Total Req, Total Current (IT), I1, I2, and the voltage at 5 different locations (V1-5).

Answer 1

Given:

circuit look like this

Required: calculate R equivalent and current in branches

Explanation:

we can see that resistor between B and D are in series then they will add like as

$\begin{gathered} R_{BD}=R_2+R_3 \\ R_{BD}=319+300 \\ R_{BD}=619\text{ ohm} \end{gathered}$

now resistor between

$R_{BD}$

and E are in parallel combination

and we call it Resistor R

$\begin{gathered} R_{BE}=\frac{R_{BD}\times R_E}{R_{BD}+R_E} \\ R_{BE}=\frac{619\times500}{619+500} \\ R_{BE}=276.58\text{ ohm} \end{gathered}$

now this resistor is in series with 100 ohm then we have equivalent resistor is

$\begin{gathered} R_{eq}=100+276.58 \\ R_{eq}=376.58\text{ ohm} \end{gathered}$

now calculate the total current in the circuit

that is

$V=IR_{eq}$

Plugging all the values in the above relation we get

$\begin{gathered} 12\text{ V }=I\times376.58\text{ ohm} \\ I=\frac{12\text{ V}}{376.58} \\ I=0.031\text{ A} \end{gathered}$

Above is the total current flowing in the circuit.

total current will be in the circuit consist of two parts

$\begin{gathered} I=I_1+I_2......(1) \\ \end{gathered}$

now apply Kirchhoff's voltage law in the first loop and solve current 1, we get

$\begin{gathered} 12-100I_-500I_1=0 \\ I_1=\frac{12-100\times0.031}{500} \\ I_1=0.0178\text{ A} \end{gathered}$

now put the value of the above current 1 in the equation(1), we get

$\begin{gathered} I=I_1+I_2 \\ I_2=I-I_1 \\ I_2=0.031-0.0178 \\ I_2=0.0132\text{ A} \end{gathered}$

now calculate voltage the across all the resistor

voltage across resistor 1 is

$\begin{gathered} V_1=R_1I_ \\ V_1=100\times0.031 \\ V_1=3.1\text{ V} \end{gathered}$

voltage across across 500 ohm resistance

$\begin{gathered} V_2=500\times0.0178 \\ V_2=8.9\text{ V} \end{gathered}$

Voltage across 310 ohm is

$\begin{gathered} V_3=319\times0.0132 \\ V_3=4.21\text{ V} \end{gathered}$

Voltage across 300 ohm is

$\begin{gathered} V_4=300\times0.0132 \\ V_4=3.96\text{ V} \end{gathered}$