Suppose that a 9 Volt battery is connected in series with a 0.3Ohm resistor in a

Question

Suppose that a 9 Volt battery is connected in series with a 0.3Ohm resistor in a closed-loop circuit, but the battery itself has a comparable internal resistance r = 0.3 Ohm. What is the current in the loop in steady state? What is the voltage drop across the resistor? What is the voltage drop across the battery? At what rate is heat dissipated by the resistor? At what rate is heat dissipated by the battery? If the battery is well insulated, and its heat capacity is 10 J/K. about how many minutes can the circuit be closed before the temperature of the battery rises to a dangerous level, say 451 degrees F?

Explanation / Answer

Emf of battery = 9. volt

Resistance = 0.3 ohm

Internal resistance = 0.3 ohm

So equivalent resistance = 0.6 ohm

(a.). Current in the loop = 9/0.6 = 15 A

(b.). Voltage drop across resister = 15×0.3 = 4.5 volt

(c.) Heat dissipated by resistor = 152×0.3 = 67.5 watt

   HeHeat dissipated by battey wil be ssame = 67.5 watt.

(d.). 452 F = 232.78°C

10×232.78 = 67.5×time

Or. time = 34.49. seconds

I.e time taken = 0.57 minute

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