In the arrangement shown in the figure,the cu | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Using Kirchhoff's voltage law:For loop $ABCDA$,$2i_1 + 5(i_1 + i_2) = 12$,which simplifies to $7i_1 + 5i_2 = 12 \dots (1)$For loop $EBCFE$,$2i_2 +

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(A) Using Kirchhoff's voltage law:For loop $ABCDA$,$2i_1 + 5(i_1 + i_2) = 12$,which simplifies to $7i_1 + 5i_2 = 12 \dots (1)$For loop $EBCFE$,$2i_2 + 5(i_1 + i_2) = 12$,which simplifies to $5i_1 + 7i_2 = 12 \dots (2)$Now,multiplying equation $(1)$ by $7$ and equation $(2)$ by $5$ gives:$49i_1 + 35i_2 = 84 \dots (3)$$25i_1 + 35i_2 = 60 \dots (4)$Subtracting equation $(4)$ from $(3)$ gives $24i_1 = 24$,so $i_1 = 1\,A$.Substituting $i_1 = 1\,A$ into equation $(1)$,we get $7(1) + 5i_2 = 12$,so $5i_2 = 5$,which means $i_2 = 1\,A$.The current through the $5\,\Omega$ resistor is $i_1 + i_2 = 1 + 1 = 2\,A$.

In the arrangement shown in the figure,the current through the $5\,\Omega$ resistor is ............. $A$.

Similar Questions

$A$ non-zero current passes through the galvanometer $G$ shown in the circuit when the key $K$ is closed and its value does not change when the key is opened. Then,which of the following statement$(s)$ is/are true?

Five resistors,each of resistance $R$,are arranged as shown in the figure. $A$ battery of voltage $V$ is connected between points $A$ and $B$. Find the current flowing through the branch $AFCEB$.

The current $I$ in the circuit shown is:

What are the advantages of the null-point method in a Wheatstone bridge? What additional measurements would be required to calculate $R_{\text{unknown}}$ by any other method?

In the following network,$I_{1} = -0.4 \text{ A}$,$I_{4} = 1 \text{ A}$ and $I_{5} = 0.4 \text{ A}$. The values of $I_{2}$,$I_{3}$ and $I_{6}$ respectively are:

Vedclass Test Series

Exam Paper Generator

Online Exam Module