Question
Question: A wire of resistance \[9\,\Omega \] is connected with the two batteries as shown in the adjoining ci...
A wire of resistance 9Ω is connected with the two batteries as shown in the adjoining circuit. The potential difference between the points A and B is:
A. 15V
B. −15V
C. 3V
D. −3V
Solution
To find the potential difference between the points A and B, we need to find the current flowing in the circuit. Here, we will need to apply Kirchhoff’s voltage law. Using this law, we can find the value of current flowing in the circuit and using this value of current we can find the potential difference between A and B.
Complete step by step answer:
Given, resistance of wire, R=9Ω
Emf of one battery, ε1=8V and internal resistance, r1=1Ω
Emf of other battery, ε2=12V and internal resistance, r2=2Ω
Let us assume current i flows through the circuit.We draw a diagram to show the flow of current.
We will apply here Kirchhoff’s voltage law, according to which the algebraic sum of all the potential differences around any closed loop is equal to zero.The given circuit is a closed circuit, so we can use Kirchhoff’s voltage law and find the current in the circuit.
Applying Kirchhoff’s voltage law in the given circuit we get,
2×i−12+1×i+8+9×i=0
⇒2i−12+i+8+9i=0
⇒12i−4=0
⇒i=124A
Potential difference between two points is given by,
V=IR
where I is the current flowing between the two points and R is the resistance between the two points.
So, here potential difference between the points A and B will be
V=iR
Putting the value of i and R we get,
V=124×9
⇒V=31×9
∴V=3V
Therefore, the potential difference between the points A and B is 3V.
Hence, the correct answer is option C.
Note: There are two Kirchhoff’s laws. First is the Kirchhoff’s current law which states in a junction, total current entering the junction is equal to total current leaving the junction. Second is the Kirchhoff’s voltage law which we have discussed in the above question. For solving problems related to circuits, these two laws are very important.