A 9 V battery is hooked up to two resistors in series.One has a resistance of 5 Ω,and the other has a resistance of 10 Ω.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?
A)the 10 Ω resistor
B)the 5 Ω resistor
C)Energy is being dissipated by both resistors at the same rate.

Question

A 9 V battery is hooked up to two resistors in series.One has a resistance of 5 Ω,and the other has a resistance of 10 Ω.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? 
A)the 10 Ω resistor 
B)the 5 Ω resistor 
C)Energy is being dissipated by both resistors at the same rate.

Quizplus · Accepted Answer

The rate at which energy is dissipated by a resistor can be calculated using the formula P = V^2/R, where P is the power (in watts), V is the voltage across the resistor, and R is the resistance of the resistor. Since the two resistors are in series, the current through both resistors is the same, and the voltage is split between the two resistors according to their resistances. Using Ohm's law (V = IR), we can find that the voltage across the 10 Ω resistor is twice the voltage across the 5 Ω resistor. Therefore, the power dissipated by the 10 Ω resistor is four times the power dissipated by the 5 Ω resistor. Hence, the energy is being dissipated at a higher rate by the 10 Ω resistor.

A 9 V Battery Is Hooked Up to Two Resistors