Answer:
72.80 % more energy will be consumed.
Explanation:
First of all, let us have a look at the formula of energy for a processor.
[tex]E = CV^2f[/tex]
Where, [tex]E[/tex] is the energy
[tex]C[/tex] is the capacitance
[tex]V[/tex] is the voltage and
[tex]f[/tex] is the clock rate.
Let [tex]E_1[/tex] be the energy of older processor.
[tex]C_1[/tex] be the capacitance of older processor.
[tex]V_1[/tex] be the voltage of older processor
[tex]f_1[/tex] be the capacitance of older processor
So, [tex]E_1 = C_1V_1^2f_1[/tex] ....... (1)
and
Let [tex]E_2[/tex] be the energy of newer processor.
[tex]C_2[/tex] be the capacitance of newer processor.
[tex]V_2[/tex] be the voltage of newer processor
[tex]f_2[/tex] be the capacitance of newer processor
[tex]E_2 = C_2V_2^2f_2[/tex] ....... (2)
Dividing equation (2) with equation (1):
[tex]\dfrac{E_2}{E_1} = \dfrac{C_2V_2^2f_2}{C_1V_1^2f_1}[/tex]
As per given statement:
[tex]C_1=C_2[/tex]
[tex]V_2=1.2\times V_1[/tex]
[tex]f_2=1.2\times f_1[/tex]
Putting the values above:
[tex]\dfrac{E_2}{E_1} = \dfrac{C_1\times (1.2V_1)^2\times 1.2f_1}{C_1V_1^2f_1}\\\Rightarrow \dfrac{E_2}{E_1} = \dfrac{(1.2)^2\times 1.2}{1}\\\Rightarrow \dfrac{E_2}{E_1} = 1.728\\\Rightarrow \bold{E_2 = 1.728 \times E_1}[/tex]
Energy consumed by newer processor is 1.728 times the energy consumed by older processor.
OR
72.80 % more energy will be consumed by the newer processor.
