Question

In the circuit shown below, the potential difference between A and B is: graphics[width=0.5\linewidth]{Q1.png}

• A. 6 V
• B. 3 V
• C. 2 V
• D. 1 V

Hint:

For parallel cells with equal internal resistances: \[ E_{\text{eq}}=\frac{E_1+E_2+E_3+\cdots}{n} \]

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The given circuit diagram consists of multiple branches connected in parallel between two nodes, which we can label as D and C.
This is a classic problem involving the parallel combination of real voltage sources.
A real voltage source is modeled as an ideal EMF (\(E\)) in series with an internal resistance (\(r\)).
In this specific circuit, we see three such branches between points D and C.
Furthermore, point A is connected to D through a resistor, and point B is connected to C through a resistor.
Because terminals A and B are open-ended (no load is connected), no current can flow out of A or B.
According to Ohm's Law (\(V = IR\)), if the current \(I\) is zero, the voltage drop across those specific series resistors is zero.
Consequently, the potential at A is identical to the potential at D (\(V_A = V_D\)), and the potential at B is identical to the potential at C (\(V_B = V_C\)).
The problem thus asks us to find the potential difference across the parallel network of the three batteries.
Step 2: Key Formula or Approach:
To find the equivalent EMF (\(E_{eq}\)) of \(n\) batteries connected in parallel with their positive terminals on one side and negative on the other, we use Millman's Theorem or the equivalent EMF formula:
\[ E_{eq} = \frac{\frac{E_1}{r_1} + \frac{E_2}{r_2} + \dots + \frac{E_n}{r_n}}{\frac{1}{r_1} + \frac{1}{r_2} + \dots + \frac{1}{r_n}} \]
This formula accounts for the fact that each branch contributes to the total potential difference based on its strength and its internal resistance.
Step 3: Detailed Explanation:
Let's analyze the parameters for the three parallel branches:
Branch 1 (Top): EMF \(E_1 = 1 V\), Resistance \(r_1 = 1 \Omega\).
Branch 2 (Middle): EMF \(E_2 = 2 V\), Resistance \(r_2 = 1 \Omega\).
Branch 3 (Bottom): EMF \(E_3 = 3 V\), Resistance \(r_3 = 1 \Omega\).
Now, we apply the equivalent EMF formula:
\[ E_{eq} = \frac{\frac{1}{1} + \frac{2}{1} + \frac{3}{1}}{\frac{1}{1} + \frac{1}{1} + \frac{1}{1}} \]
Simplifying the numerator: \(1 + 2 + 3 = 6\).
Simplifying the denominator: \(1 + 1 + 1 = 3\).
Therefore, \[ E_{eq} = \frac{6}{3} = 2 V. \]
The equivalent internal resistance (\(r_{eq}\)) of the parallel combination would be:
\[ \frac{1}{r_{eq}} = \frac{1}{1} + \frac{1}{1} + \frac{1}{1} = 3 \implies r_{eq} = \frac{1}{3} \Omega. \]
The potential difference across nodes D and C is \(V_D - V_C = E_{eq} = 2 V\).
As discussed in Step 1, since no current flows through the external resistors connected to A and B, there is no potential drop across them.
\(V_A = V_D\) and \(V_B = V_C\).
Thus, the potential difference between A and B is:
\[ V_{AB} = V_A - V_B = V_D - V_C = 2 V. \]
Step 4: Final Answer:
The potential difference between A and B is calculated to be 2 V.