A 9-V battery is hooked up to two resistors in series. One has a resistance of 5
ID: 1398504 • Letter: A
Question
A 9-V battery is hooked up to two resistors in series. One has a resistance of 5 ?, and the other has a resistance of 10 ?. Several locations along the circuit are marked with letters, as shown in the figure. Through which resistor is energy being dissipated at the higher rate?
Item 1 Part A A 9-V battery is hooked up to two resistors in series. One has a resistance of 5 , and the other has a resistance of 10 . Several locations along the circuit are marked with letters, as shown in the figure. Through which resistor is energy being dissipated at the higher rate? 5 9 VT 10 O the 10-2 resistor O Energy is being dissipated by both resistors at the same rate. the 5- resistorExplanation / Answer
Rate of dissipation of energy is power.
Power= i^2* R
So first we need to find current i in the circuit.
i = Vnet /Rnet
=9 / (10+5)
=0.6 A
power for 5 ohm = i^2 * 5
= (0.6^2) * 5
= 1.8 W
power for 10 ohm = i^2 * 10
= (0.6^2) * 10
= 3.6 W
So, 10 ohm resistor would dissipate at higher rate
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