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Step-by-Step Solution
Step 1: Represent the Battery and Resistances
Let the electromotive force (emf) of the battery be $ \varepsilon $ and the internal resistance of the battery be $ r $. When a resistor $R_1 = 2\,\Omega$ is connected across the battery, the current flowing is $ I_1 = 2\,\text{A} $. When a resistor $R_2 = 9\,\Omega$ is connected, the current flowing is $ I_2 = 0.5\,\text{A} $.
Step 2: Write the Current Equation for the First Case
Using Ohm’s law and taking into account the internal resistance $ r $, the net resistance in the circuit is $ (R_1 + r) $. The current is given by:
$ I_1 = \frac{\varepsilon}{R_1 + r} $
Substituting the given values ($ I_1 = 2\,\text{A} $ and $ R_1 = 2\,\Omega $), we get:
$ 2 = \frac{\varepsilon}{2 + r} \quad \dots (1)$
Step 3: Write the Current Equation for the Second Case
Similarly, for the second resistor $ R_2 = 9\,\Omega $, the total resistance is $ (9 + r) $, and the current is $ I_2 = 0.5\,\text{A} $. Thus:
$ I_2 = \frac{\varepsilon}{R_2 + r} $
Substituting the given values ($ I_2 = 0.5\,\text{A} $ and $ R_2 = 9\,\Omega $), we get:
$ 0.5 = \frac{\varepsilon}{9 + r} \quad \dots (2)$
Step 4: Divide the Two Equations to Eliminate $ \varepsilon $
Dividing equation (1) by equation (2) gives:
$ \frac{I_1}{I_2} = \frac{\frac{\varepsilon}{2 + r}}{\frac{\varepsilon}{9 + r}} = \frac{9 + r}{2 + r} $
Using the given currents $ I_1 = 2\,\text{A} $ and $ I_2 = 0.5\,\text{A} $:
$ \frac{2}{0.5} = \frac{9 + r}{2 + r} $
Hence:
$ 4 = \frac{9 + r}{2 + r} $
Step 5: Solve for $ r $
Cross-multiplying:
$ 4(2 + r) = 9 + r $
$ 8 + 4r = 9 + r $
$ 4r - r = 9 - 8 $
$ 3r = 1 $
$ r = \frac{1}{3}\,\Omega $
Final Answer
The internal resistance of the battery is $ \frac{1}{3}\,\Omega $.
