A 9 V battery is connected in series with a resistor .The terminal vol

1.	A 9 V battery is connected in series with a resistor .The terminal voltage is found to be 8 V. Current through the circuit is measured as 5 A. What is the internal resistance of the battery?
Answer» r = ε - V/I = 9V - 8V/5A = 0.2 Ω

A 9 V battery is connected in series with a resistor .The terminal voltage is found to be 8 V. Current through the circuit is measured as 5 A. What is the internal resistance of the battery?

Answer»

r = ε - V/I

= 9V - 8V/5A

= 0.2 Ω

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A 9 V battery is connected in series with a resistor .The terminal voltage is found to be 8 V. Current through the circuit is measured as 5 A. What is the internal resistance of the battery?

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