Let's solve the problem step by step. **Given:** - Three resistors: \( R_{ab} = 15.0\,\Omega \), \( R_{bc} = 10.0\,\Omega \), \( R_{ca} = 20.0\,\Omega \) - Internal resistance of battery: \( r = 3.0\,\Omega \) - The battery is connected across \( bc \) (so current flows from \( b \) to \( c \)) - We are to find the **current** the array draws (usually this means with a certain EMF, but let's assume \( V = 1\,\text{V} \) for calculation, since no voltage is given; in such problems, you can express the answer in terms of \( V \) if needed). --- ### **Step 1: Find the Equivalent Resistance Across \( b \) and \( c \)** - Between \( b \) and \( c \): a \( 10\,\Omega \) resistor. - There is also a path \( b \to a \to c \) (through \( 15\,\Omega \) and \( 20\,\Omega \)). These two paths are **in parallel**: \[ R_{bac} = R_{ba} + R_{ac} = 15\,\Omega + 20\,\Omega = 35\,\Omega \] So, the total resistance between \( b \) and \( c \) is: \[ \frac{1}{R_{eq}} = \frac{1}{10} + \frac{1}{35} \] \[ \frac{1}{R_{eq}} = \frac{35 + 10}{350} = \frac{45}{350} = \frac{9}{70} \] \[ R_{eq} = \frac{70}{9} \approx 7.78\,\Omega \] **Explanation:** We combine the two parallel paths (\( 10\,\Omega \) and \( 35\,\Omega \)) using the parallel formula. --- ### **Step 2: Add the Internal Resistance of the Battery** The total resistance in the circuit is: \[ R_{total} = R_{eq} + r = 7.78\,\Omega + 3.0\,\Omega = 10.78\,\Omega \] **Explanation:** The internal resistance of the battery is in series with the external (array) resistance. --- ### **Step 3: Calculate the Current (Assuming a Voltage \( V \))** Using Ohm's Law: \[ I = \frac{V}{R_{total}} \] If the battery voltage \( V \) is not given, leave your answer in terms of \( V \): \[ I = \frac{V}{10.78} \] If \( V = 1\,\text{V} \), then: \[ I = \frac{1}{10.78} \approx 0.093\,\text{A} \] **Explanation:** Use Ohm's Law with the total resistance to find the current. --- ## **Final Answer** **If the battery voltage is \( V \), the current drawn is:** \[ \boxed{I = \frac{V}{10.78}\ \text{A}} \] **For a 1 V battery,** \[ \boxed{I \approx 0.093~\text{A}} \]

The problem involves analyzing a resistor network connected to a battery with internal resistance. The key concepts used are combining resistors in parallel to find the equivalent resistance between two points, adding internal resistance in series to determine total circuit resistance, and then applying Ohm's Law to calculate the current based on the battery voltage. These steps allow us to simplify the complex resistor arrangement into a single equivalent resistance and compute the current drawn by the circuit.