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Silver and copper voltmeters are connect...

Silver and copper voltmeters are connected in parallel with a battery of emf `12V`. In `30` minutes, `1g` of silver and `1.8g` of copper are liberated. The power supplied by the battery is
`(Z_(Cu) = 6.6xx10^(-4) g//C and Z_(Ag) = 11.2xx10^(-4)g//C`)

A

720 J

B

2.41 J

C

24.12 J

D

`4.34xx10^4J`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to determine the power supplied by the battery when silver and copper are liberated in the given amounts. We will use the formula for current and the relationship between power, current, and voltage. ### Step-by-Step Solution: 1. **Understanding the Problem**: We have a battery with an emf of 12V, and two voltmeters (one for silver and one for copper) connected in parallel. In 30 minutes, 1g of silver and 1.8g of copper are liberated. 2. **Convert Time to Seconds**: \[ T = 30 \text{ minutes} = 30 \times 60 = 1800 \text{ seconds} \] 3. **Calculate the Current for Silver**: The current \( I_{Ag} \) due to the liberation of silver can be calculated using the formula: \[ I_{Ag} = \frac{M_{Ag}}{Z_{Ag} \cdot T} \] where: - \( M_{Ag} = 1 \text{ g} \) - \( Z_{Ag} = 11.2 \times 10^{-4} \text{ g/C} \) - \( T = 1800 \text{ s} \) Substituting the values: \[ I_{Ag} = \frac{1}{11.2 \times 10^{-4} \cdot 1800} \] 4. **Calculate the Current for Copper**: Similarly, the current \( I_{Cu} \) for copper is: \[ I_{Cu} = \frac{M_{Cu}}{Z_{Cu} \cdot T} \] where: - \( M_{Cu} = 1.8 \text{ g} \) - \( Z_{Cu} = 6.6 \times 10^{-4} \text{ g/C} \) Substituting the values: \[ I_{Cu} = \frac{1.8}{6.6 \times 10^{-4} \cdot 1800} \] 5. **Total Current**: The total current \( I \) supplied by the battery is the sum of the currents through silver and copper: \[ I = I_{Ag} + I_{Cu} \] 6. **Calculate Power Supplied by the Battery**: The power \( P \) supplied by the battery can be calculated using the formula: \[ P = V \cdot I \] where \( V = 12 \text{ V} \). 7. **Substituting Values**: After calculating \( I_{Ag} \) and \( I_{Cu} \), substitute these values into the equation for power: \[ P = 12 \cdot (I_{Ag} + I_{Cu}) \] 8. **Final Calculation**: Perform the calculations to find the total power supplied by the battery.
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