Answer
405.3k+ views
Hint: As a convention, for voltage sources, when moving from negative to positive add voltage when moving from positive to negative subtract voltage. For resistors, when moving in the same direction as conventional current, subtract voltage across the resistor but when moving in the opposite direction add voltage across the resistor.
Formula used: $\sum V = 0$ where $V$ is the voltage around a loop, $\sum {{I_{in}} = \sum {{I_{out}}} } $ where ${I_{in}}$ is the current flowing into a node and ${I_{out}}$ is the current flowing out of the node.
Complete step by step answer
To solve, let us apply Kirchhoff’s voltage rule around the loop BAFEB, we have
$\Rightarrow 9 - {I_2}(1) - {I_1}(1) - {I_2} - 6 = 0 $
$\Rightarrow 9 - 6 = 2{I_1} + {I_2} = 3$
Since ${I_2} = 0$, then
$\Rightarrow {I_1} = \dfrac{3}{2} $
$\Rightarrow 1.5A $
At node E,
$\Rightarrow {I_1} = {I_2} + {I_3} $
$\Rightarrow {I_1} = {I_3} $
Since ${I_2} = 0$, using the Kirchhoff’s voltage law on loop BAFDCB we get
$\Rightarrow 9 - {I_1} - {I_1} - {I_1}R - 3 = 0 $
$\Rightarrow 6 - 2{I_1} = {I_1}R $
Making $R$ subject of the formula and inserting the values for ${I_1}$ we have
$\Rightarrow R = \dfrac{{6 - 2\left( {1.5} \right)}}{{1.5}} $
$\Rightarrow \dfrac{3}{{1.5}} $
$\Rightarrow 2\Omega $
The voltage between A and D is simply the voltage remained if we apply Kirchhoff’s law around a loop DCAFD but stop at point A. i.e.
$\Rightarrow {V_{DA}} = - {I_1}R - 3 + 9 = - {I_1}R + 6$
Inputting values of ${I_1}$ and $R$ again and solving, we get
$\Rightarrow {V_{DA}} = - 1.5\left( 2 \right) + 6 $
$\Rightarrow - 3 + 6 $
$\Rightarrow {V_{DA}} = 3V$
Hence, the required answer is 3V.
Additional Information
Kirchhoff’s rule is a physical principle which is more fundamental than Ohm’s law as it was derived from the principle of conservation of charge. In fact, it works with other elements besides resistors such as capacitors and inductors. It can also be used to analyze low frequency ac current.
Note
Alternatively, to find ${V_{DA}}$ we can take the path DFA instead of the path DCA as done above. Thus, we have,
$\Rightarrow {V_{DA}} = {I_1} + {I_1} = 2{I_1}$
The voltage difference across the $1\Omega $ resistors because we move in the direction opposite to the assumed conventional flow.
Substituting the value of ${I_1}$ in the above equation we get,
$\Rightarrow {V_{DA}} = 2\left( {1.5} \right) = 3V$
$\therefore {V_{DA}} = 3V$
Formula used: $\sum V = 0$ where $V$ is the voltage around a loop, $\sum {{I_{in}} = \sum {{I_{out}}} } $ where ${I_{in}}$ is the current flowing into a node and ${I_{out}}$ is the current flowing out of the node.
Complete step by step answer
![seo images](https://www.vedantu.com/question-sets/1bfd48ee-7f62-4cf8-8e18-93eccba7046d8454919694374376266.png)
To solve, let us apply Kirchhoff’s voltage rule around the loop BAFEB, we have
$\Rightarrow 9 - {I_2}(1) - {I_1}(1) - {I_2} - 6 = 0 $
$\Rightarrow 9 - 6 = 2{I_1} + {I_2} = 3$
Since ${I_2} = 0$, then
$\Rightarrow {I_1} = \dfrac{3}{2} $
$\Rightarrow 1.5A $
At node E,
$\Rightarrow {I_1} = {I_2} + {I_3} $
$\Rightarrow {I_1} = {I_3} $
Since ${I_2} = 0$, using the Kirchhoff’s voltage law on loop BAFDCB we get
$\Rightarrow 9 - {I_1} - {I_1} - {I_1}R - 3 = 0 $
$\Rightarrow 6 - 2{I_1} = {I_1}R $
Making $R$ subject of the formula and inserting the values for ${I_1}$ we have
$\Rightarrow R = \dfrac{{6 - 2\left( {1.5} \right)}}{{1.5}} $
$\Rightarrow \dfrac{3}{{1.5}} $
$\Rightarrow 2\Omega $
The voltage between A and D is simply the voltage remained if we apply Kirchhoff’s law around a loop DCAFD but stop at point A. i.e.
$\Rightarrow {V_{DA}} = - {I_1}R - 3 + 9 = - {I_1}R + 6$
Inputting values of ${I_1}$ and $R$ again and solving, we get
$\Rightarrow {V_{DA}} = - 1.5\left( 2 \right) + 6 $
$\Rightarrow - 3 + 6 $
$\Rightarrow {V_{DA}} = 3V$
Hence, the required answer is 3V.
Additional Information
Kirchhoff’s rule is a physical principle which is more fundamental than Ohm’s law as it was derived from the principle of conservation of charge. In fact, it works with other elements besides resistors such as capacitors and inductors. It can also be used to analyze low frequency ac current.
Note
Alternatively, to find ${V_{DA}}$ we can take the path DFA instead of the path DCA as done above. Thus, we have,
$\Rightarrow {V_{DA}} = {I_1} + {I_1} = 2{I_1}$
The voltage difference across the $1\Omega $ resistors because we move in the direction opposite to the assumed conventional flow.
Substituting the value of ${I_1}$ in the above equation we get,
$\Rightarrow {V_{DA}} = 2\left( {1.5} \right) = 3V$
$\therefore {V_{DA}} = 3V$
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)