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.Which resistor is converting electric energy into internal energy 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.Which resistor is converting electric energy into internal energy 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 power dissipated by a resistor can be determined using the formula P=V^2/R, where P is power, V is voltage, and R is resistance. Since both resistors are in series with the battery, the current passing through both resistors is the same. Therefore, the power dissipated by each resistor is proportional to its resistance. The 10-Ω resistor has a higher resistance than the 5-Ω resistor, so it is converting electric energy into internal energy at a higher rate.

A 9-V Battery Is Hooked Up to Two Resistors in Series.One