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Step-by-Step Solution
Step 1: Identify the Known Quantities
• EMF of the battery, $E = 3.0\text{ V}$
• Terminal voltage of the battery, $V = 2.5\text{ V}$
• Power dissipated in the external resistor, $P_R = 0.5\text{ W}$
• We need to find the power dissipated in the internal resistance, $P_{\text{internal}}$.
Step 2: Relate Power to Current and Resistance
Power dissipated by the external resistor $R$ is given by:
$$P_R = i^2 R.$$
We know $P_R = 0.5\text{ W}$, but we do not directly know $i$ or $R$. However, the terminal voltage across the resistor is $2.5\text{ V}$, so we can also write:
$$P_R = i \times V_{\text{across }R} = i \times 2.5.$$
Both relations will help find $i$ and $R$.
Step 3: Calculate the Current
From the definition of power across $R$:
$$P_R = i \times 2.5 = 0.5\text{ W}.$$
Hence,
$$i = \frac{0.5}{2.5} = 0.2\text{ A}.$$
So the current in the circuit is $0.2\text{ A}$.
Step 4: Determine the Resistance of the External Resistor
Using $P_R = i^2 R$, we have:
$$0.5 = (0.2)^2 \times R.$$
So,
$$R = \frac{0.5}{(0.2)^2} = \frac{0.5}{0.04} = 12.5\,\Omega.$
Step 5: Find the Internal Resistance of the Battery
The terminal voltage $V$ is given by
$$V = E - i\,r,$$
where $r$ is the internal resistance of the battery. Substituting the known values:
$$2.5 = 3.0 - (0.2)\,r.$$
So,
$$r = \frac{3.0 - 2.5}{0.2} = \frac{0.5}{0.2} = 2.5\,\Omega.$
Step 6: Calculate the Power Dissipated in the Internal Resistance
The power dissipated in the internal resistance is:
$$P_{\text{internal}} = i^2\,r = (0.2)^2 \times 2.5 = 0.04 \times 2.5 = 0.1\text{ W}.$$
Thus, the power dissipated internally is $0.10\text{ W}$.
