Question

For a cell or a battery, the emf is (A) equal to the potential difference between its terminals when terminals are not connected externally (B) less than the potential difference between its terminals when the cell/battery is being discharged (C) always greater than the potential difference between its terminals (D) less than the potential difference between its terminals when the cell/battery is being charged Choose the correct answer from the options given below:

• A. (A) and (D) only
• B. (A) and (C) only
• C. (C) only
• D. (A), (B) and (C) only

Hint:

Think of internal resistance as something that always opposes the direction current is trying to flow through the cell. Work out separately what happens to terminal voltage when current leaves the cell during discharge versus when current is forced into the cell during charging, and remember the special case where no current flows at all.

Answer 1

The Correct Option is A

Concept: For a cell of emf \(E\), terminal potential difference \(V\), current \(I\), and internal resistance \(r\): During discharging, \[ V=E-Ir \] During charging, \[ V=E+Ir \]

Step 1: Check statement (A). When the cell is not connected externally, \[ I=0 \] Therefore, \[ V=E \] Hence, \[ \boxed{\text{Statement (A) is true.}} \]

Step 2: Check statement (B). During discharging, \[ V=E-Ir \] which implies \[ V<E \] or \[ E>V \] Thus emf is greater than the terminal potential difference. Hence, \[ \boxed{\text{Statement (B) is false.}} \]

Step 3: Check statement (C). When discharging, \[ E>V \] but when the cell is open, \[ E=V \] Therefore emf is not always greater than terminal voltage. Hence, \[ \boxed{\text{Statement (C) is false.}} \]

Step 4: Check statement (D). During charging, \[ V=E+Ir \] Therefore, \[ V>E \] or \[ E<V \] Hence emf is less than the terminal potential difference. \[ \boxed{\text{Statement (D) is true.}} \]

Step 5: State the answer. \[ \boxed{ \text{Statements (A) and (D) are correct.} } \] Hence, the correct option is \[ \boxed{(A)} \]

Answer 2

Concept:

• Internal resistance can be treated as a component that always opposes the direction of current flowing through the cell itself, so tracing the circuit loop directly (Kirchhoff Voltage Law) gives the terminal voltage in any situation without needing to memorise two separate formulas.
• Terminal voltage $V$ is what is actually available to the external circuit; emf $E$ is the full voltage the cell is capable of producing internally.

Step 1: Apply the loop rule for the discharging case, where the cell drives current through an external circuit.
Current $I$ leaves the positive terminal, passes through the internal resistance $r$, and some voltage is used up in that direction of flow. Tracing the loop gives $V = E - Ir$, so $V < E$ while the cell is discharging.

Step 2: Apply the loop rule for the charging case, where an external source drives current into the cell against its own emf.
Now current $I$ is forced in the reverse direction, so the external source must overcome both the emf and the internal resistance drop together. Tracing the loop gives $V = E + Ir$, so $V > E$ while the cell is charging.

Step 3: Apply the loop rule for the open-circuit case.
With no external circuit connected, $I = 0$. Both expressions above collapse to $V = E$, since there is no current left to create any drop across $r$.

Step 4: Test statement (A) using Step 3, and statement (D) using Step 2.
Statement (A) claims $V = E$ when the terminals are not connected externally, which matches Step 3 exactly - TRUE. Statement (D) claims emf is less than terminal voltage while charging, i.e. $E < V$, which matches Step 2 exactly - TRUE.

Step 5: Test statement (B) using Step 1, and statement (C) using Step 3.
Statement (B) claims emf is less than terminal voltage while discharging, i.e. $E < V$, but Step 1 gives $V = E - Ir$, meaning $E > V$ instead - FALSE. Statement (C) claims emf is always greater than terminal voltage, but Step 3 shows they become equal at open circuit - FALSE.

Final Answer: (A) and (D) only