A voltmeter is attached to the terminals of an “unloaded” or standalone 9 V batt

Question

A voltmeter is attached to the terminals of an “unloaded” or standalone 9 V battery. The meter reads 9.4 V. What is the EMF of this battery? What is the terminal voltage of this battery? How much current is flowing through the battery?

in the previous question, a 200 resistor is placed across the terminals of the 9 V battery. A voltmeter reading across the battery terminals shows a voltage of 8.8 V. How much current is flowing through the resistor? How much current is flowing through the battery? calculate the internal resistance of the battery.

Explanation / Answer

V = terminal potential difference = 9.4 Volts

E = emf = terminal potential difference = 9.4 Volts

since the battery is unloaded , i = 0 A

terminal Voltage is given as

V = 8.8 volts

R = 200 ohm

current is given as

i = V/R = 8.8/200 = 0.044 A

same current flows through the battery

V = E - ir

8.8 = 9 - (0.044) r

r = 4.5 ohm

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