Two batteries of having emfs E1 and E2 and internal resistances r1 and r2 respectively are joined as shown. VA and VB are the potentials at A and B respectively.

Let E2>E1, then I=E2−E1r1+r2For potential difference across battery 1, apply KVL to A1B, we get VA−E1−Ir1−VB=0⇒VA−VB=E1+Ir1 (1)⇒VA−VB=E1+E2−E1r1+r2r1 ⇒VA−VB=E1r1+E1r2+E2r1−E1r1r1+r2⇒VA−VB=E1r2+E2r1r1+r2 (2)For potential difference across battery 2, apply KVL to A2B, we getVA+Ir2−E2−VB=0⇒VA−VB=E2−Ir2 (3)⇒VA−VB=E2−E2−E1r1+r2r2⇒VA−VB=E2r1+E2r2−E2r2+E1r2r1+r2⇒ VA−VB=E1r2+E2r1r1+r2 (4) Also, from equation (1), we get that potential difference across 1 is greater than its emf E1.Now, since the current I flows from the positive plate to the negative plate inside the battery 1, hence energy is absorbed by it or in other words the battery will continuously get the energy supplied by the other battery. Hence all the statements 1, 2, 3 and 4 are correct.

Two batteries of having emfs E1 and E2 and internal resistances r1 and r2 respectively are joined as shown. VA and VB are the potentials at A and B respectively.

Let E2>E1, then I=E2−E1r1+r2For potential difference across battery 1, apply KVL to A1B, we get VA−E1−Ir1−VB=0⇒VA−VB=E1+Ir1 (1)⇒VA−VB=E1+E2−E1r1+r2r1 ⇒VA−VB=E1r1+E1r2+E2r1−E1r1r1+r2⇒VA−VB=E1r2+E2r1r1+r2 (2)For potential difference across battery 2, apply KVL to A2B, we getVA+Ir2−E2−VB=0⇒VA−VB=E2−Ir2 (3)⇒VA−VB=E2−E2−E1r1+r2r2⇒VA−VB=E2r1+E2r2−E2r2+E1r2r1+r2⇒ VA−VB=E1r2+E2r1r1+r2 (4) Also, from equation (1), we get that potential difference across 1 is greater than its emf E1.Now, since the current I flows from the positive plate to the negative plate inside the battery 1, hence energy is absorbed by it or in other words the battery will continuously get the energy supplied by the other battery. Hence all the statements 1, 2, 3 and 4 are correct.

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Two batteries of having emfs E₁ and E₂ and internal resistances r₁ and r₂ respectively are joined as shown. V_A and V_B are the potentials at A and B respectively.

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One battery will continuously supply energy to the other.

The potential difference across one battery will be greater than its emf.

The potential difference across both the battery will be equal.

$V_{A} - V_{B} = \frac{(E_{1} r_{2} + E_{2} r_{1})}{(r_{1} + r_{2})}$

answer is A, B, C, D.

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Detailed Solution

$\begin{array}{r} Let E_{2} > E_{1}, then \\ I = \frac{E_{2} - E_{1}}{r_{1} + r_{2}} \end{array}$

For potential difference across battery 1, apply KVL to A1B, we get

$\begin{array}{l} V_{A} - E_{1} - {Ir}_{1} - V_{B} = 0 \\ \Rightarrow V_{A} - V_{B} = E_{1} + {Ir}_{1} (1) \\ \Rightarrow V_{A} - V_{B} = E_{1} + (\frac{E_{2} - E_{1}}{r_{1} + r_{2}}) r_{1} \end{array}$

$\begin{array}{l} \Rightarrow V_{A} - V_{B} = \frac{E_{1} r_{1} + E_{1} r_{2} + E_{2} r_{1} - E_{1} r_{1}}{r_{1} + r_{2}} \\ \Rightarrow V_{A} - V_{B} = \frac{E_{1} r_{2} + E_{2} r_{1}}{r_{1} + r_{2}} (2) \end{array}$

For potential difference across battery 2, apply KVL to A2B, we get

$\begin{array}{l} V_{A} + {Ir}_{2} - E_{2} - V_{B} = 0 \\ \Rightarrow V_{A} - V_{B} = E_{2} - {Ir}_{2} (3) \\ \Rightarrow V_{A} - V_{B} = E_{2} - (\frac{E_{2} - E_{1}}{r_{1} + r_{2}}) r_{2} \\ \Rightarrow V_{A} - V_{B} = \frac{E_{2} r_{1} + E_{2} r_{2} - E_{2} r_{2} + E_{1} r_{2}}{r_{1} + r_{2}} \end{array}$

$\Rightarrow V_{A} - V_{B} = \frac{E_{1} r_{2} + E_{2} r_{1}}{r_{1} + r_{2}} (4)$

Also, from equation (1), we get that potential difference across 1 is greater than its emf E₁.

Now, since the current I flows from the positive plate to the negative plate inside the battery 1, hence energy is absorbed by it or in other words the battery will continuously get the energy supplied by the other battery. Hence all the statements 1, 2, 3 and 4 are correct.

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