Calculate Terminal Voltage Across Two Parallel Batteries

To find the terminal voltage across $AB$ ($V_{AB} = V_A - V_B$), we analyze the given parallel combination of two real batteries.

1. Circuit Parameters

From the circuit diagram:

• Top branch:
  • EMF, $E_1 = 6\text{ V}$
  • Internal resistance, $r_1 = 1\ \Omega$
• Bottom branch:
  • EMF, $E_2 = 3\text{ V}$
  • Internal resistance, $r_2 = 2\ \Omega$

Both batteries are connected with their positive terminals towards terminal $A$ and negative terminals towards terminal $B$.

---

2. Derivation using Nodal Analysis

Let the potential at terminal $B$ be reference zero, i.e., $V_B = 0\text{ V}$.
Let the potential at terminal $A$ be $V_A = V$.

Since the terminals $A$ and $B$ are in an open circuit (no current leaves terminal $A$ or enters terminal $B$ externally), the sum of currents leaving node $A$ through both branches must equal zero by Kirchhoff's Current Law (KCL):

$\frac{V - E_1}{r_1} + \frac{V - E_2}{r_2} = 0$

Substitute the given values into the equation:

$\frac{V - 6}{1} + \frac{V - 3}{2} = 0$

Multiply the entire equation by $2$ to clear the denominator:

$2(V - 6) + (V - 3) = 0$

$2V - 12 + V - 3 = 0$

$3V - 15 = 0$

$3V = 15 \implies V = 5\text{ V}$

---

3. Equivalent EMF Formula Method

Alternatively, using the standard formula for equivalent EMF ($E_{\text{eq}}$) of cells connected in parallel:

$V_{AB} = E_{\text{eq}} = \frac{\frac{E_1}{r_1} + \frac{E_2}{r_2}}{\frac{1}{r_1} + \frac{1}{r_2}}$

Substituting the values:

$V_{AB} = \frac{\frac{6}{1} + \frac{3}{2}}{\frac{1}{1} + \frac{1}{2}} = \frac{6 + 1.5}{1 + 0.5} = \frac{7.5}{1.5} = 5\text{ V}$

Thus, the voltage across $AB$ is $5\text{ V}$.