Question

A battery of emf 10 V and internal resistance 3 Ohm is connected to a resistor. If the current in the circuit is 0.5 A, what is the resistance of the resistor? What is the terminal voltage of the battery when the circuit is closed?

Answer 1

Given Information:

• EMF of the battery (E): 10 V
• Internal resistance (r): 3 Ω
• Current in the circuit (I): 0.5 A

---

1. What is the resistance of the resistor (R)?

To find the resistance of the external resistor, we use the formula for a complete circuit which relates EMF, current, and total resistance. The total resistance is the sum of the external resistance (R) and the internal resistance (r).

Formula:
$E = I \times (R + r)$

Now, we can plug in the values we know and solve for R:

1. $10 \, \text{V} = 0.5 \, \text{A} \times (R + 3 \, \Omega)$
2. Divide both sides by 0.5 A:
   $\frac{10}{0.5} = R + 3$
3. $20 = R + 3$
4. Subtract 3 from both sides to find R:
   $R = 20 - 3$
   $R = 17 \, \Omega$

Answer: The resistance of the resistor is 17 Ω.

---

2. What is the terminal voltage of the battery (V_T)?

The terminal voltage is the potential difference across the battery's terminals when the circuit is closed and current is flowing. It is the EMF minus the voltage "lost" due to the internal resistance.

Formula:
$V_T = E - I \times r$

Let's plug in the values:

1. $V_T = 10 \, \text{V} - (0.5 \, \text{A} \times 3 \, \Omega)$
2. $V_T = 10 \, \text{V} - 1.5 \, \text{V}$
3. $V_T = 8.5 \, \text{V}$

Alternatively, the terminal voltage is also equal to the voltage drop across the external resistor (R). We can use this to check our answer.
$V_T = I \times R$
$V_T = 0.5 \, \text{A} \times 17 \, \Omega$
$V_T = 8.5 \, \text{V}$

Both methods give the same result.

Answer: The terminal voltage of the battery is 8.5 V.