4.2.3 Electromotive force and potential difference (3)
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1.
A battery with a potential difference of 12 V is connected to a circuit containing a resistor and a light bulb. The current flowing through the circuit is measured to be 0.5 A. Calculate the power supplied by the battery.
Given:
- Potential difference (V) = 12 V
- Current (I) = 0.5 A
Equation: Power (P) = V x I
Calculation:
P = 12 V x 0.5 A = 6 W
Answer: The power supplied by the battery is 6 W.
2.
A battery with a voltage of 9V is connected to a resistor. A current of 0.5 A flows through the resistor. Calculate the potential difference (p.d.) across the resistor.
Answer:
We know that the potential difference (p.d.) across a resistor is related to the voltage (V) and the current (I) by Ohm's Law: V = IR.
Given: V = 9V, I = 0.5 A
Therefore, V = 0.5 A * 9 V = 4.5 V
The potential difference across the resistor is 4.5V.
3.
A student is using a voltmeter to measure the potential difference across a cell. The voltmeter is connected in parallel with the cell. The voltmeter reads 2.5V. What does this reading indicate about the cell?
Answer:
The voltmeter reading of 2.5V indicates the potential difference (p.d.) between the two terminals of the cell. This is the electrical potential difference that exists across the cell's terminals when it is connected in a circuit. It shows the 'push' or 'force' that drives the current through the circuit. A reading of 2.5V means that there is a potential difference of 2.5 volts between the positive and negative terminals of the cell.