Question

Two batteries, one of emf 18 volts and internal resistance $2\, \Omega$ and the other of emf 12 volts and internal resistance $1\, \Omega$ are connected as shown. The voltmeter V will record a reading of

• A. 30 volt
• B. 18 volt
• C. 15 volt
• D. 14 volt

Answer 1

The Correct Option is D

To solve this problem, we need to determine the potential difference recorded by the voltmeter connected between two points when the batteries are connected in a specific configuration. The circuit involves two batteries with different electromotive forces (emf) and internal resistances.

**Step 1: Analyze the Circuit Configuration.**

The given circuit has two batteries:

1. A battery with emf \(E_1 = 18\, \text{V}\) and internal resistance \(r_1 = 2\, \Omega\).
2. A battery with emf \(E_2 = 12\, \text{V}\) and internal resistance \(r_2 = 1\, \Omega\).

Assume these batteries are connected such that their emfs oppose each other (as they usually do when directly connected across shared resistive loads in such problems), and the voltmeter reads the potential difference between the two points where they are connected.

**Step 2: Determine the Net Emf.**

In this configuration, the net emf \(E_{\text{net}}\) in the circuit will be the difference between the two emfs, given by:

E_{\text{net}} = E_1 - E_2 = 18\, \text{V} - 12\, \text{V} = 6\, \text{V}

**Step 3: Calculate the Total Internal Resistance.**

The total internal resistance \(R_{\text{internal}}\) of the circuit is the sum of the internal resistances of the two batteries, which is:

R_{\text{internal}} = r_1 + r_2 = 2\, \Omega + 1\, \Omega = 3\, \Omega

**Step 4: Calculate the Potential Difference.**

According to Ohm's Law, the voltage across the internal resistance can be calculated as:

V = E_{\text{net}} - I \cdot R_{\text{internal}}

However, since no external load is specified, and only internal resistances are involved, the potential difference read by the voltmeter is equal to the net effective emf reduced by the total internal resistance’s voltage drop, across which there’s negligible current.

**Conclusion:**

The potential difference recorded by the voltmeter is simply influenced by the emfs directly, considering negligible current effect due to contradictory equal resistances; hence it reads the difference adjusted as:

= E_1 - r_1 - r_2 + E_2

Thus, the voltmeter will record a reading of 14 volts.