Djd hdjf jdjfjfjfj jdjfjfjfj jdjfjfjfj jdjfjfjfj jdjfjfjfj jdjfjfjfj
Answer:
Explanation:
For ligation process the 1:3 vector to insert ratio is the good to utilize . By considering that we can take 1 ratio of vector and 3 ratio of insert ( consider different insert size ) and take 10 different vials of ligation ( each calculated using different insert size from low to high ) and plot a graph for transformation efficiency and using optimum transformation efficiency we can find out the insert size.
Answer:
The upper diode conduces in the odd half cycles. The lower diode conduces in the even half cycles.
Explanation:
The peak voltage after the 6 to 1 step down is . Then, the peak voltage of the rectified output is V_{d}[/tex] and according to the statement, the diodes can be modeled to be . Then, the peak voltage in the load is .
The upper diode conduces in the odd half cycles. The lower diode conduces in the even half cycles.
The average output voltage is calculated as:
The average current in the load is calculated as:
Answer:
Explanation:
the solution to the problem is given in the pictures attached. (b) is answered first then (a). I hope the explanation helps you.Thank you
Answer:
first step here is to substitute the 3 of your two equations into the second;
3 Ne^(-Q_v/k(1293)) = Ne^(-Q_v/k(1566))
Since 'N' is a constant, we can remove it from both sides.
We also want to combine our two Q_v values, so we can solve for Q_v, so we should put them both on the same side:
3 = e^(-Q_v/k(1293)) / e^(-Q_v/k(1566))
3 = e^(-Q_v/k(1293) + Q_v/k(1566) ) (index laws)
ln (3) = -Q_v/k(1293) + Q_v/k(1566) (log laws)
ln (3) = -0.13Q_v / k(1566) (addition of fractions)
Q_v = ln (3)* k * 1566 / -0.13 (rearranging the equation)
Now, as long as you know Boltzmann's constant it's just a matter of substituting it for k and plugging everything into a calculator.