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A battery is made by joining m rows of i...

A battery is made by joining m rows of identical cells in parallel. Each row consists of n cells joined in series. This battery sends a maximum current I in a given external resistor. Now the cells are so areaged that instead of m rows. N rows are joind in parallel and each row consists of m cells joined in series. Find the current through the same external resistor (Total number of cells which is equal to nm is connected )

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To solve the problem, we need to analyze the two configurations of the battery and how they affect the current flowing through the external resistor. ### Step-by-Step Solution: 1. **Understanding the Initial Configuration**: - Initially, we have **m rows** of cells in parallel, with each row containing **n cells** in series. - The total number of cells in this configuration is \( N = m \times n \). - The maximum current sent through the external resistor is given as \( I \). 2. **Voltage and Resistance in the Initial Configuration**: - Let the voltage of each cell be \( E \). - The total voltage across each row (which is in series) is \( V_{\text{initial}} = nE \). - The equivalent resistance of each row (assuming each cell has a resistance \( r \)) is \( R_{\text{row}} = nr \). - The total resistance of the battery when m rows are in parallel is given by: \[ R_{\text{total, initial}} = \frac{R_{\text{row}}}{m} = \frac{nr}{m} \] 3. **Applying Ohm's Law**: - The current \( I \) through the external resistor \( R \) can be expressed using Ohm's law: \[ I = \frac{V_{\text{initial}}}{R_{\text{total, initial}} + R} \] - Substituting the values: \[ I = \frac{nE}{\frac{nr}{m} + R} \] 4. **Understanding the New Configuration**: - In the new configuration, we have **N rows** in parallel, with each row containing **m cells** in series. - The total number of cells remains \( N = m \times n \). - The total voltage across each row is now \( V_{\text{new}} = mE \). - The equivalent resistance of each row is \( R_{\text{row, new}} = mr \). - The total resistance of the battery when N rows are in parallel is: \[ R_{\text{total, new}} = \frac{R_{\text{row, new}}}{N} = \frac{mr}{N} = \frac{mr}{n} \] 5. **Applying Ohm's Law in the New Configuration**: - The new current \( I' \) through the external resistor can be expressed as: \[ I' = \frac{V_{\text{new}}}{R_{\text{total, new}} + R} \] - Substituting the values: \[ I' = \frac{mE}{\frac{mr}{n} + R} \] 6. **Finding the Relationship Between Currents**: - To find the ratio of the new current \( I' \) to the original current \( I \): \[ \frac{I'}{I} = \frac{\frac{mE}{\frac{mr}{n} + R}}{\frac{nE}{\frac{nr}{m} + R}} \] - Simplifying this expression will give us the relationship between the two currents. 7. **Final Expression**: - After simplification, we can find the new current \( I' \) in terms of the original current \( I \): \[ I' = \frac{m}{n} \cdot I \] ### Final Answer: The current through the same external resistor in the new configuration is given by: \[ I' = \frac{m}{n} \cdot I \]
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