Answer
64.8k+ views
Hint: We will apply here Kirchhoff’s second law, i.e., $\sum ir = \sum e$. It is an application of mesh analysis. After analyzing the loops one by one by, we have to solve the two equations to get the value of the current.
Formula used:
let, the current from the battery in the loop is $i$ and resistance of the loop is $r$. Then the product of these two will be the electromotive force acting in that loop, i.e., $\sum ir = \sum e$.
Complete step by step solution:
Kirchhoff formulated two laws. One is Kirchhoff’s first law or Kirchhoff’s current law (KCL). It states that in an electrical circuit (or network of wires) the algebraic sum of currents through the conductors meeting at a point is zero, i.e., $\sum i = 0$.
Another one is Kirchhoff’s second law or Kirchhoff’s voltage law (KVL). It states that the algebraic sum of the product of the current and resistance in any closed loop of a circuit is equal to the algebraic sum of electromotive force acting in that loop, i.e., $\sum ir = \sum e$
The given circuit is a mesh, a complicated circuit formed by a number of adjacent loops. So, it can be analyzed one by one by applying Kirchhoff’s second law.
Let, current ${i_1}$ flow in the 1st mesh and current ${i_2}$ flow in the 2nd mesh.
Now, applying Kirchhoff’s second law in the mesh 1, we get,
$2{i_1} + 6{i_1} - 6{i_2} + 8{i_1} + 0.5{i_1} = 15$
or, $16.5{i_1} - 6{i_2} = 15$ $ \ldots \left( 1 \right)$
Applying Kirchhoff’s second law in the mesh 2, we get,
$7{i_2} + {i_2} + 10{i_2} + 6{i_2} - 6{i_1} = 0$
or, $24{i_2} - 6{i_1} = 0$
or, $4{i_2} - {i_1} = 0$ $ \ldots \left( 2 \right)$
From equation $\left( 2 \right)$ we get, ${i_1} = 4{i_2}$
Putting the value of ${i_1}$ in the equation $\left( 1 \right)$, we get,
$16.5 \times 4{i_2} - 6{i_2} = 15$
or, $66{i_2} - 6{i_2} = 15$
or, $60{i_2} = 15$
or, ${i_2} = \dfrac{{15}}{{60}} = \dfrac{1}{4}$
Now, putting this value of ${i_2}$ in equation $\left( 2 \right)$, we get,
$(4 \times \dfrac{1}{4}) - {i_1} = 0$
or, ${i_1} = 1$
So, the current from the battery in the circuit is $1A$.
Note: While applying this law, we have to keep in mind the sign convention. In traversing a loop, the currents flowing in the clockwise direction are taken as positive while currents flowing in the anticlockwise direction are taken as negative. Also, while traversing the loop, the electromotive force of the source, which sends current in the clockwise direction, is taken as positive and the emf of the source, which sends current in the anticlockwise direction, is negative.
Formula used:
let, the current from the battery in the loop is $i$ and resistance of the loop is $r$. Then the product of these two will be the electromotive force acting in that loop, i.e., $\sum ir = \sum e$.
Complete step by step solution:
Kirchhoff formulated two laws. One is Kirchhoff’s first law or Kirchhoff’s current law (KCL). It states that in an electrical circuit (or network of wires) the algebraic sum of currents through the conductors meeting at a point is zero, i.e., $\sum i = 0$.
Another one is Kirchhoff’s second law or Kirchhoff’s voltage law (KVL). It states that the algebraic sum of the product of the current and resistance in any closed loop of a circuit is equal to the algebraic sum of electromotive force acting in that loop, i.e., $\sum ir = \sum e$
The given circuit is a mesh, a complicated circuit formed by a number of adjacent loops. So, it can be analyzed one by one by applying Kirchhoff’s second law.
Let, current ${i_1}$ flow in the 1st mesh and current ${i_2}$ flow in the 2nd mesh.
Now, applying Kirchhoff’s second law in the mesh 1, we get,
$2{i_1} + 6{i_1} - 6{i_2} + 8{i_1} + 0.5{i_1} = 15$
or, $16.5{i_1} - 6{i_2} = 15$ $ \ldots \left( 1 \right)$
Applying Kirchhoff’s second law in the mesh 2, we get,
$7{i_2} + {i_2} + 10{i_2} + 6{i_2} - 6{i_1} = 0$
or, $24{i_2} - 6{i_1} = 0$
or, $4{i_2} - {i_1} = 0$ $ \ldots \left( 2 \right)$
From equation $\left( 2 \right)$ we get, ${i_1} = 4{i_2}$
Putting the value of ${i_1}$ in the equation $\left( 1 \right)$, we get,
$16.5 \times 4{i_2} - 6{i_2} = 15$
or, $66{i_2} - 6{i_2} = 15$
or, $60{i_2} = 15$
or, ${i_2} = \dfrac{{15}}{{60}} = \dfrac{1}{4}$
Now, putting this value of ${i_2}$ in equation $\left( 2 \right)$, we get,
$(4 \times \dfrac{1}{4}) - {i_1} = 0$
or, ${i_1} = 1$
So, the current from the battery in the circuit is $1A$.
Note: While applying this law, we have to keep in mind the sign convention. In traversing a loop, the currents flowing in the clockwise direction are taken as positive while currents flowing in the anticlockwise direction are taken as negative. Also, while traversing the loop, the electromotive force of the source, which sends current in the clockwise direction, is taken as positive and the emf of the source, which sends current in the anticlockwise direction, is negative.
Recently Updated Pages
Write a composition in approximately 450 500 words class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Arrange the sentences P Q R between S1 and S5 such class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What is the common property of the oxides CONO and class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What happens when dilute hydrochloric acid is added class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
If four points A63B 35C4 2 and Dx3x are given in such class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
The area of square inscribed in a circle of diameter class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Other Pages
A boat takes 2 hours to go 8 km and come back to a class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Electric field due to uniformly charged sphere class 12 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
In the ground state an element has 13 electrons in class 11 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
According to classical free electron theory A There class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Differentiate between homogeneous and heterogeneous class 12 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Excluding stoppages the speed of a bus is 54 kmph and class 11 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)