Question

The value of current I for the circuit below is:
Given,
R=1kohm
E=12V

*it is a request to kindly explain your answer..the process through which you are solving this problem*

Find the circuit diagram attached.

1. 1.9 mA
2. 2.9 mA
3. 3.9 mA
4. 4.9mA

Answer 1

Solution :-

**refer to attached for the reference.

First of all we will add some more information in the diagram in order to solve it.

As In AMG , A battery is connected we will take a current $i_1$

Then according to Kirchoff's junction law at junction G.

$i_1 = I + x$

$x = i_1 - I$

So current in GB = $i_1 - I$

Now in FED we will take another current $i_2$ .

$I = i_2 + y$

$y = I - i_2$

So current in FC = $I - i_2$

Now by applying junction law at other junctions

▪️Current in CB = $I$

▪️Current in BA = $i_1$

Now we will write Loop For (anticlockwise)

(1)AHGB

$-12 = 3R(i_1 - I) + R(i_1)$

$\implies -12 = 3Ri_1 - 3RI + Ri_1$

$\implies -12 = 4Ri_1 - 3RI$...(i)

(2) AHFC

$-12 = 4R(I - i_2) + R(i_1)$

$\implies -12 = 4RI - 4Ri_2 + Ri_1$...(ii)

(3) AHED

$24 - 12 = 2R(i_2) + R(i_1)$

$\implies12 = 2Ri_2 + Ri_1$ ....(iii)

(4)CFED

$24 = 2R(i_2) - 4R(I-i_2)$

$\implies 24 = 2Ri_2 - 4RI + 4Ri_2$

$\implies 24 = 6Ri_2 - 4RI$ ....(iv)

Now 3(iv) - 4(i)

$72 = 18Ri_2 - 12RI$

$+ 48 = - 16Ri_1 + 12RI$

________________________

$120 = 18Ri_2 - 16Ri_1$ .....(v)

________________________

Now from (v) - 9(iii)

$120 = 18Ri_2 - 16Ri_1$

$-108= - 18Ri_2 - 9Ri_1$

________________________

$12 = -25Ri_1$

________________________

$Ri_1 = \dfrac{-12}{25}$

Now from by substiting value of $Ri_1$ in (i)

$-12 = 4Ri_1 - 3RI$

$\implies -12 = 4\dfrac{-12}{25} - 3RI$

$\implies -12 = \dfrac{-48}{25}- 3RI$

$\implies 3RI = \dfrac{-48}{25} + 12$

$\implies 3RI = \dfrac{300-48}{25}$

$\implies 3RI = \dfrac{252}{25}$

$\implies 3RI = 10.08$

$\implies RI = 3.36$

$\implies I = \dfrac{3.36}{R}$

$\implies I = \dfrac{3.36}{1000}$

$\implies I = 3.36 \: mA$

So current I = 3.36 mA

Answer 2

Solution :-

**refer to attached for the reference.

First of all we will add some more information in the diagram in order to solve it.

As In AMG , A battery is connected we will take a current

Then according to Kirchoff's junction law at junction G.

So current in GB =

Now in FED we will take another current .

So current in FC =

Now by applying junction law at other junctions

▪️Current in CB =

▪️Current in BA =

Now we will write Loop For (anticlockwise)

(1)AHGB

...(i)

(2) AHFC

...(ii)

(3) AHED

....(iii)

(4)CFED

....(iv)

Now 3(iv) - 4(i)

________________________

.....(v)

________________________

Now from (v) - 9(iii)

________________________

________________________

Now from by substiting value of in (i)

So current I = 3.36 mA