Question

A battery of emf \(15\ \text{V}\) and internal resistance of \(4\ \Omega\) is connected to a resistor. If the current in the circuit is \(2\ \text{A}\), the resistance of the resistor and terminal voltage of the battery will be:

• A. \(2.5\ \Omega,\ 6\ \text{V}\)
• B. \(3.5\ \Omega,\ 6\ \text{V}\)
• C. \(2.5\ \Omega,\ 7\ \text{V}\)
• D. \(3.5\ \Omega,\ 7\ \text{V}\)

Hint:

Remember: \[ E=I(R+r) \] and \[ V=E-Ir \]

• Terminal voltage decreases when current flows
• Internal resistance consumes part of emf

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
A real battery contains an internal resistance ($r$) within its chemical cells. Because of this, when the battery drives an electrical current ($I$) through an external circuit, a portion of its electromotive force ($\text{emf}, E$) is dropped internally across this resistance. The actual voltage available at the external output terminals is called the terminal voltage ($V$), and it is always lower than the ideal $\text{emf}$ value when current is flowing.
Step 2: Key Formula or Approach:
1. Total current in a single-loop closed circuit using Ohm's Law: \[ I = \frac{E}{R + r} \] 2. Terminal voltage equation during circuit discharge: \[ V = E - Ir \quad \text{or} \quad V = IR \] Where $E = 15 \text{ V}$, $r = 4 \ \Omega$, and $I = 2 \text{ A}$.
Step 3: Detailed Explanation:
First, let's determine the external resistance value ($R$) using our loop current formula: \[ I = \frac{E}{R + r} \implies 2 \text{ A} = \frac{15 \text{ V}}{R + 4 \ \Omega} \] Cross-multiply to clear the fraction: \[ 2(R + 4) = 15 \] \[ 2R + 8 = 15 \] \[ 2R = 15 - 8 = 7 \] \[ R = \frac{7}{2} = 3.5 \ \Omega \] Next, compute the terminal voltage ($V$) of the battery: Using the discharge formula: \[ V = E - Ir = 15 \text{ V} - (2 \text{ A} \times 4 \ \Omega) \] \[ V = 15 - 8 = 7 \text{ V} \] Alternatively, you can verify this using Ohm's law across the external resistor: \[ V = IR = 2 \text{ A} \times 3.5 \ \Omega = 7 \text{ V} \] Both calculation paths confirm that the resistance is $3.5 \ \Omega$ and the terminal voltage is $7\text{ V}$. This matches option (D).
Step 4: Final Answer:
The resistance of the external resistor is $3.5 \ \Omega$ and the terminal voltage of the battery is $7\text{ V}$.