Question

Find the source voltage across the 1 mA current source.

Answer 1

Answer:

R equivalent of large triangle is 16x8/24= 16/3

total resistance is 1+16/3=19/3

voltage is = IxRnet= 6.3×10^-3

Answer 2

Concept

Resistance is the opposition that a substance offers to the flow of electric current.

Given

The current passes through the resistors shown in the figure $1mA$.

Find

We have to find the source voltage

Solution

Firstly, we have to combine $6k$ and $2k$ series resistors i.e. $R_{1}=6k+2k=8k$, then add the value of the series equivalent resistor to the value of the  $8k$ resistor in the original circuit which is in parallel. So,

$\frac{1}{R_{2}}=\frac{1}{8}+\frac{1}{8}$

$\Rightarrow R_{2}=4$.

We don't have any parallel or series reductions at this time.

Now, we can combine $R_{1},R_{2}$ with resistor of $R_{3}=4K$ which is passing from the center of both.

$R_{4}=\frac{R_{1}R_{2}+R_{2}R_{3}+R_{3}R_{1}}{R}$

$R_{4}=\frac{16k+16k+16k}{4k}$

$R_{4}=12k$

Now, we have two pair of resistors that we combine in parallel

$\frac{1}{R_{5}}=\frac{1}{12}+\frac{1}{3}$

$R_{5}=3$

Now, we can solve the resistor $3k,3k,12k$ which all are in parallel

$R_{6}=\frac{36}{18}$

$\Rightarrow R_{6}=2$

Similarly, $R_{7}=2k$

Now, we will find $R_{8}$ with the help of $3k,3k$, we get

$R_{8}=\frac{9}{18}$

$R_{8}=0.5$

Now, we will combine $4k$ with $R_{8}$ which is in series

$R_{9}=4+0.5=4.5k$

Similarly, we will combine $2k$ and $4k,$ we get

$R_{10}=2+4=6k$

Here, $R_{9}$ and $R_{10}$ are in parallel;

$R_{11}=\frac{1}{4.5}+\frac{1}{6}$

$R_{11}=2.5k$

At last, all the resistors we have left is series resistors

$R_{12}=1k+2k+2.5k=5.5k$

So, to find the voltage across the current source, we use the current value and the resistor value to find the voltage across the equivalent resistor is

$1mA\times5.5=5.5V$

Hence, the source voltage is $5.5V$.

#SPJ2