In the given circuit, calculate the potential difference between points A and B.
Answer
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Hint: Firstly assign arbitrary directions for current flowing through every resistor. Now recall Kirchhoff's voltage rule or the loop rule which says that the voltage in a closed loop to be zero and apply the same to the two loops under consideration. Then solve for the currents and accordingly get the answer.
Complete answer:
In the question we are given a circuit which is a combination of 3 cells and 3 resistors and we are supposed to find the potential difference between the points A and B.
Firstly, let us mark the current directions.
Now, let us apply Kirchoff’s voltage rule in loop 1.
$-3{{I}_{1}}+10-4+2{{I}_{2}}=0$
$\Rightarrow 2{{I}_{2}}-3{{I}_{1}}=-6$ ……………………………………………… (1)
Now on applying kirchhoff's voltage rule in loop 2, we get,
$-2{{I}_{2}}+4-2-\left( {{I}_{1}}+{{I}_{2}} \right)=0$
$\Rightarrow 3{{I}_{2}}+{{I}_{1}}=2$ ……………………………………………. (2)
Now we have to solve the equations (1) and (2) in two variables. Let us multiply equation (2) with 3 to get,
$9{{I}_{2}}+3{{I}_{1}}=6$ …………………………………. (3)
Now, let add (1) and (3) to get,
${{I}_{2}}=0A$
Substituting this in equation (2) we get,
${{I}_{1}}=2A$
Now we know that the potential difference across nodes A and B would be the sum of the voltage across the resistor and the cell.
${{V}_{AB}}={{V}_{A}}-{{V}_{B}}$
$\Rightarrow {{V}_{AB}}=-4+2{{I}_{2}}=-4+2\left( 0 \right)$
$\therefore {{V}_{AB}}=-4V$
Therefore, we found the potential difference across points A and B to be -4V.
Note:
Since the given combination consists of parallel connections, we could conclude that the potential difference remains the same across all the three lines though we have found it using the middle one. Also, we have considered the assigned sign by assuming the direction opposite to that of current flow to be negative throughout the solution.
Complete answer:
In the question we are given a circuit which is a combination of 3 cells and 3 resistors and we are supposed to find the potential difference between the points A and B.
Firstly, let us mark the current directions.
Now, let us apply Kirchoff’s voltage rule in loop 1.
$-3{{I}_{1}}+10-4+2{{I}_{2}}=0$
$\Rightarrow 2{{I}_{2}}-3{{I}_{1}}=-6$ ……………………………………………… (1)
Now on applying kirchhoff's voltage rule in loop 2, we get,
$-2{{I}_{2}}+4-2-\left( {{I}_{1}}+{{I}_{2}} \right)=0$
$\Rightarrow 3{{I}_{2}}+{{I}_{1}}=2$ ……………………………………………. (2)
Now we have to solve the equations (1) and (2) in two variables. Let us multiply equation (2) with 3 to get,
$9{{I}_{2}}+3{{I}_{1}}=6$ …………………………………. (3)
Now, let add (1) and (3) to get,
${{I}_{2}}=0A$
Substituting this in equation (2) we get,
${{I}_{1}}=2A$
Now we know that the potential difference across nodes A and B would be the sum of the voltage across the resistor and the cell.
${{V}_{AB}}={{V}_{A}}-{{V}_{B}}$
$\Rightarrow {{V}_{AB}}=-4+2{{I}_{2}}=-4+2\left( 0 \right)$
$\therefore {{V}_{AB}}=-4V$
Therefore, we found the potential difference across points A and B to be -4V.
Note:
Since the given combination consists of parallel connections, we could conclude that the potential difference remains the same across all the three lines though we have found it using the middle one. Also, we have considered the assigned sign by assuming the direction opposite to that of current flow to be negative throughout the solution.
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