Hi Mr Shone, I’m not very sure about this question. Why is it that the current flowing through will be halved? Wouldn’t the power of the kettle stay the same and P=IV, so when V is halved, I will be doubled instead? Thank you!
The resistance is assumed to be the same. Thus when the voltage halves the current will halve and the power will thus be one quarter of the initial value. (From P=IV or P=I²R)
Remember “220 V, 2000 W” means that 220 V is the operating voltage and that the power consumed by the device will be 2000 Wwhen it is connected to a 220 V supply.
Hi Mr Shone, I’m not very sure about this question. Why is it that the current flowing through will be halved? Wouldn’t the power of the kettle stay the same and P=IV, so when V is halved, I will be doubled instead? Thank you!
No, power does not stay the same!
The resistance is assumed to be the same. Thus when the voltage halves the current will halve and the power will thus be one quarter of the initial value. (From P=IV or P=I²R)
Remember “220 V, 2000 W” means that 220 V is the operating voltage and that the power consumed by the device will be 2000 W when it is connected to a 220 V supply.