For an ideal transformer power loss is assumed to be zero
i.e. the power in primary coil due to input voltage must be equal to power in secondary coil due to output voltage
this can be written in form of equation
here we know that
So primary coil voltage is 49 Volts
The capacitance of a capacitor is the ratio of the stored charge to its potential difference, i.e.
C = Q/ΔV
C is the capacitance
Q is the stored charge
ΔV is the potential difference
Rearrange the equation:
ΔV = Q/C
We also know the capacitance of a parallel-plate capacitor is given by:
C = κε₀A/d
C is the capacitance
κ is the capacitor's dielectric constant
ε₀ is the electric constant
A is the area of the plates
d is the plate separation
If we substitute C:
ΔV = Qd/(κε₀A)
We assume the stored charge and the area of the plates don't change. Then if we double the plate spacing, i.e. we double the value of d, then the potential difference ΔV is also doubled.
If she walks 132 and 6 you do 132 x 6 = 792
Answer:
The COP of the system is = 4.6
Explanation:
Given data
Higher pressure = 1.8 M pa
Lower pressure = 0.12 M pa
Now we have to find out high & ow temperatures at these pressure limits.
Higher temperature corresponding to pressure 1.8 M pa
°c = 335.9 K
Lower temperature corresponding to pressure 0.2 M pa
°c = 262.9 K
COP of the system is given by
COP = 4.6
Therefore the COP of the system is = 4.6