Q.A storage battery of emf $8.0\ \text{V}$ and internal resistance $0.5\ \Omega$ is being charged by a $120\ \text{V}$ dc supply using a series resistor of $15.5\ \Omega$. What is the terminal voltage of the battery during charging? What is the purpose of having a series resistor in the charging circuit?

Concept: Internal Resistance – During charging, the battery’s terminal voltage is not its emf; it is the emf plus the voltage drop across its internal resistance because the charging current flows into the positive terminal.

Reasoning:

1. Net voltage in the circuit: supply voltage minus battery emf (which opposes charging).

   $V_{\text{net}} = 120\ \text{V} - 8.0\ \text{V} = 112\ \text{V}$.

2. Total resistance in series: $R_{\text{ext}} + r = 15.5\ \Omega + 0.5\ \Omega = 16.0\ \Omega$.

3. Charging current:

   $I = \frac{V_{\text{net}}}{R_{\text{total}}} = \frac{112}{16.0} = 7.0\ \text{A}$. …

1. Understanding the situation — charging a battery

When a battery is being charged, current is forced into its positive terminal. This is the opposite of discharging, where current flows out of the positive terminal.

The key difference:

• Discharging: terminal voltage $V = \varepsilon - I r$ (emf minus internal drop)
• Charging: terminal voltage $V = \varepsilon + I r$ (emf plus internal drop)

Why? Because the internal resistance $r$ opposes the flow of current. During charging, the external supply pushes current against the battery's emf, so the internal resistance causes an additional voltage rise across the battery terminals.

2. Finding the charging current

The circuit: a $120\ \text{V}$ DC supply, a series resistor $R = 15.5\ \Omega$, and the battery (emf $\varepsilon = 8.0\ \text{V}$, internal resistance $r = 0.5\ \Omega$) all in series.

The net voltage driving current is the supply voltage minus the battery's back emf (since the battery opposes the charging current):

The total resistance in the circuit is:

So the charging current is:

3. Terminal voltage during charging

The terminal voltage of the battery is the voltage measured across its terminals. During charging:

Substitute:

$V_{\text{terminal}} = 8.0\ \text{V} + (7.0\ \text{A})(0.5\ \Omega) = 8.0\ \text{V} + 3.5\ \text{V} = 11.5\ \text{V}$ …

Method: Kirchhoff’s Voltage Law (KVL) for Charging Circuits

When a battery is being charged, current flows into its positive terminal — opposite to discharge. This means the internal resistance causes a voltage rise across it (opposing the charging current), so the terminal voltage becomes:

Steps

1. Draw the circuit and assign current direction

   • The $120\ \text{V}$ supply pushes current through the series resistor ($15.5\ \Omega$) into the battery’s positive terminal.
   • Current $I$ flows clockwise: supply → series resistor → battery (positive to negative) → back to supply.

2. Apply KVL around the loop

   Starting from the negative terminal of the supply, going clockwise:

(The battery’s emf opposes the charging current, so it appears as a drop.)

3. Solve for current $I$

4. Find terminal voltage during charging Using the charging formula:

--- …

Common Mistakes Students Make on This Internal Resistance Problem

Mistake 1: Using the Wrong Sign for Internal Resistance During Charging

The error: Students treat the internal resistance drop as $Ir$ (subtracting it from emf), just like in a discharging circuit. They write:

This is wrong for charging.

Why it's wrong: During charging, current flows into the battery's positive terminal. The internal resistance still opposes current flow, so the potential drop across $r$ is $+Ir$ (adds to the emf). The correct relation is:

How to avoid: Always ask: Is the battery supplying energy (discharging) or receiving energy (charging)?

• Discharging: $V = \varepsilon - Ir$
• Charging: $V = \varepsilon + Ir$

Mistake 2: Forgetting to Find the Current First

The error: Students plug numbers directly into $V = \varepsilon + Ir$ without calculating $I$ from the full circuit.

Why it's wrong: The current $I$ depends on the total resistance in the loop — the external resistor and the internal resistance.

Correct approach: Apply Kirchhoff's voltage law to the entire charging loop:

Solve for $I$:

How to avoid: Draw the circuit. Label all resistances in series. Write one loop equation before touching the terminal voltage formula.

Mistake 3: Confusing Terminal Voltage with Emf

The error: Students answer $V = 8.0\ \text{V}$ (the emf), thinking terminal voltage equals emf during charging.

Why it's wrong: Terminal voltage is the voltage across the battery terminals — which includes the $Ir$ drop. During charging, this is greater than the emf.

Correct calculation:

How to avoid: Remember: Terminal voltage is what a voltmeter reads across the battery. During charging, it's higher than emf; during discharging, it's lower.

Mistake 4: Giving a Vague or Incorrect Purpose for the Series Resistor

The error: Students say "to limit current" without explaining why limiting is necessary, or they say "to increase voltage." …