Answer
64.8k+ views
Hint: The currents that pass through each resistor are marked and their values found out. The entire circuit is then divided into two loops. The question can be solved by applying Kirchhoff’s Voltage Law to the loops formed.
Complete Step by Step Solution:
Kirchhoff's voltage law (KVL) states that the sum of all voltages around any closed loop in a circuit must be equal to zero. This is a consequence of charge conservation and also conservation of energy. This means that the sum of all potential differences across the component involved in the circuit gives a zero reading, as expected.
We redraw the diagram as shown below.
![](https://www.vedantu.com/question-sets/9c2e7f75-c36a-41b1-8898-87f8e7174c434737676447442242964.png)
Thus,${i_1}$ is the current generated by the source of 10V and ${i_2}$ is the current generated by the 20V source.
The current passing through the resistor $1\Omega $ is the sum of ${i_1}$ and ${i_2}$ currents. Similarly, the current through the resistor $4\Omega $ is of value ${i_2}$, and the value of current passing through the $2\Omega $ resistor is of value ${i_1}$. This has been shown in the figure.
Thus there are two loops formed. Let us call the lower loop, the one with source of 20V as Loop 1 and the loop with the source 10V as Loop 2.
We apply Kirchhoff’s voltage Law to both the loops.
In the first loop of 20V, we have,
$1\left( {{i_1} + {i_2}} \right) + 4{i_2} - 20 = 0$
$ \Rightarrow 5{i_2} + {i_1} = 20$.
By applying KVL in loop 2 of 10V, we get,
$\left( {{i_1} + {i_2}} \right) + 2{i_1} - 10 = 0$
$ \Rightarrow 3{i_1} + {i_2} = 10$
From both the equations, we can write,
$5{i_2} + {i_1} = 20$ and
$2\left( {3{i_1} + {i_2}} \right) = 2 \times 10 = 20$.
Thus,
$5{i_2} + {i_1} = 6{i_1} + 2{i_2}$
$ \Rightarrow 3{i_2} = 5{i_1}$
Again, from the equations of the loops it can be written that,
$14{i_1} = 30$
$ \Rightarrow {i_1} = \dfrac{{15}}{7} = 2.1A$
The value of current through the $2\Omega $ resistor is given by the value of ${i_1}$.
Thus, the correct answer is Option D.
Note: For the second loop, where the source is of emf 20V, we have current ${i_2}$.
We know that, ${i_2} = \dfrac{5}{3}{i_1}$ and ${i_1} = \dfrac{{15}}{7}$. Thus,
${i_2} = \dfrac{5}{3} \times \dfrac{{15}}{7} = \dfrac{{25}}{7} = 3.571A$
Complete Step by Step Solution:
Kirchhoff's voltage law (KVL) states that the sum of all voltages around any closed loop in a circuit must be equal to zero. This is a consequence of charge conservation and also conservation of energy. This means that the sum of all potential differences across the component involved in the circuit gives a zero reading, as expected.
We redraw the diagram as shown below.
![](https://www.vedantu.com/question-sets/9c2e7f75-c36a-41b1-8898-87f8e7174c434737676447442242964.png)
Thus,${i_1}$ is the current generated by the source of 10V and ${i_2}$ is the current generated by the 20V source.
The current passing through the resistor $1\Omega $ is the sum of ${i_1}$ and ${i_2}$ currents. Similarly, the current through the resistor $4\Omega $ is of value ${i_2}$, and the value of current passing through the $2\Omega $ resistor is of value ${i_1}$. This has been shown in the figure.
Thus there are two loops formed. Let us call the lower loop, the one with source of 20V as Loop 1 and the loop with the source 10V as Loop 2.
We apply Kirchhoff’s voltage Law to both the loops.
In the first loop of 20V, we have,
$1\left( {{i_1} + {i_2}} \right) + 4{i_2} - 20 = 0$
$ \Rightarrow 5{i_2} + {i_1} = 20$.
By applying KVL in loop 2 of 10V, we get,
$\left( {{i_1} + {i_2}} \right) + 2{i_1} - 10 = 0$
$ \Rightarrow 3{i_1} + {i_2} = 10$
From both the equations, we can write,
$5{i_2} + {i_1} = 20$ and
$2\left( {3{i_1} + {i_2}} \right) = 2 \times 10 = 20$.
Thus,
$5{i_2} + {i_1} = 6{i_1} + 2{i_2}$
$ \Rightarrow 3{i_2} = 5{i_1}$
Again, from the equations of the loops it can be written that,
$14{i_1} = 30$
$ \Rightarrow {i_1} = \dfrac{{15}}{7} = 2.1A$
The value of current through the $2\Omega $ resistor is given by the value of ${i_1}$.
Thus, the correct answer is Option D.
Note: For the second loop, where the source is of emf 20V, we have current ${i_2}$.
We know that, ${i_2} = \dfrac{5}{3}{i_1}$ and ${i_1} = \dfrac{{15}}{7}$. Thus,
${i_2} = \dfrac{5}{3} \times \dfrac{{15}}{7} = \dfrac{{25}}{7} = 3.571A$
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)