Question

A DC supply of \(160\,V\) is used to charge a battery of EMF \(10\,V\) and internal resistance \(1\,\Omega\) by connecting a series resistance of \(24\,\Omega\). The terminal voltage of the battery during charging is:

• A. \(8 \operatorname{V} \)
• B. \(12 \operatorname{V} \)
• C. \(16 \operatorname{V} \)
• D. \(4 \operatorname{V} \)

Hint:

When a battery is being charged: 1. The external source voltage is greater than the battery's emf. 2. The current flows into the positive terminal of the battery. 3. The terminal voltage of the battery is greater than its emf: \(V_{terminal} = E + Ir\). This is because the supply must overcome the battery's emf and the voltage drop across its internal resistance. When a battery is discharging: 1. The terminal voltage is less than its emf: \(V_{terminal} = E - Ir\).

Answer 1

The Correct Option is C

Step 1: Identify the components and set up the circuit for charging.
We have a DC supply voltage \(V_{supply} = 160 \operatorname{V}\). A battery is being charged, with its own electromotive force (emf) \(E_{batt} = 10 \operatorname{V}\) and internal resistance \(r_{batt} = 1 \Omega\). An external series resistance \(R_{ext} = 24 \Omega\) is also connected. During charging, the DC supply voltage is the primary source, and it drives current against the emf of the battery being charged. The internal resistance of the battery and the external series resistance both oppose the current flow. The total effective resistance in the circuit is the sum of these resistances. 
Step 2: Calculate the total equivalent resistance of the circuit.
The external resistance and the internal resistance of the battery are in series: \[ R_{total} = R_{ext} + r_{batt} = 24 \Omega + 1 \Omega = 25 \Omega \] Step 3: Calculate the net voltage driving the current in the circuit.
The supply voltage is trying to push the current, while the battery's emf acts in opposition (it's being charged, so it's a "back emf" in this context). The net voltage available to drive the current through the total resistance is: \[ V_{net} = V_{supply} - E_{batt} = 160 \operatorname{V} - 10 \operatorname{V} = 150 \operatorname{V} \] Step 4: Calculate the charging current (\(I\)) flowing in the circuit.
Using Ohm's law, \(I = \frac{V_{net}}{R_{total}}\): \[ I = \frac{150 \operatorname{V}}{25 \Omega} = 6 \operatorname{A} \] Step 5: Calculate the terminal voltage of the battery during charging. The terminal voltage (\(V_{terminal}\)) of a battery during charging is the sum of its emf and the voltage drop across its internal resistance, because the external source needs to overcome both the battery's emf and the voltage drop across its internal resistance to push current into it: \[ V_{terminal} = E_{batt} + I r_{batt} \] \[ V_{terminal} = 10 \operatorname{V} + (6 \operatorname{A})(1 \Omega) \] \[ V_{terminal} = 10 \operatorname{V} + 6 \operatorname{V} \] \[ V_{terminal} = 16 \operatorname{V} \] The final answer is $\boxed{16 \operatorname{V}}$.