Part of the circuit for your phone charger contains a somewhat delicate resistor (which follows Ohm's law well). The resistor can only dissipate up to 0.25 Watts of power before it overheats and melts. While your charger is operating normally, the resistor typically has a current of 0.018A through it, and dissipates only 0.085 Watts of power. a) What is the resistance of this resistor? b) How much voltage is across it normally? b) What is the maximum current you can put through it before it overheats and melts?
Part of the circuit for your phone charger contains a somewhat delicate resistor (which follows Ohm's law well). The resistor can only dissipate up to 0.25 Watts of power before it overheats and melts. While your charger is operating normally, the resistor typically has a current of 0.018A through it, and dissipates only 0.085 Watts of power. a) What is the resistance of this resistor? b) How much voltage is across it normally? b) What is the maximum current you can put through it before it overheats and melts?
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Author:Raymond A. Serway, Chris Vuille
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Part of the circuit for your phone charger contains a somewhat delicate resistor (which follows Ohm's law well). The resistor can only dissipate up to 0.25 Watts of power before it overheats and melts.
While your charger is operating normally, the resistor typically has a current of 0.018A through it, and dissipates only 0.085 Watts of power.
a) What is the resistance of this resistor?
b) How much voltage is across it normally?
b) What is the maximum current you can put through it before it overheats and melts?
Expert Solution
Step 1
If current through a resistor of resistance R is I then, power dissipation by the resistor is P= I2R
Voltage across the resistor is equal to IR
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