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Hi can someone help me with this question please!
It's a passage but I will type out the part that we need:
they gave us a circuit with a power supply- a resistor R1 and a capacitor with a solution of cell suspended in a solution known as a dielectric constant: they are talking about electroporation
If the electric field is strong enough then the cells undergo dielectric breakdown opening micropores in the cell membranes. dielectric breakdown seperates the charges of a dipole momentarily so that ions fill in the space in between the plates of the capacitor. The capacitor now discharges through the cells and the cells act as resistors to the current flow. During discharge there is likely problem that cell lysis can occur. The rate of lysing can be expressed as:
L= 1-(t/tc)exp(-(E-Ec)/K)
where tc= treshold value of treatment time
t= actual treatment time
Ec= treshold Efield strength
E= actual Efield strength
K= constant
QUESTION: CELL LYSIS IS AFFECTED BY WHICH OF THE FOLLOWING FACTORS
I. AN INCREASE IN CAPACITOR PLATE AREA
II. A DECREASE IN PLATE SEPERATION
III. INCREASING THE VALUE OF R1
IV. THE NUMBER OF CELLS BEING TREATED
ANSWER is all I,II,IV
I picked all of them and don't understand why R1 doesn't affect discharge. TBR explanation is charging R1 would affect only how long it takes the capacitor to charge up. it has no effect on discharging across the capacitor and the passage states that discharging the capacitor is what sends a current through the cells that may cause them to lyse
It's a passage but I will type out the part that we need:
they gave us a circuit with a power supply- a resistor R1 and a capacitor with a solution of cell suspended in a solution known as a dielectric constant: they are talking about electroporation
If the electric field is strong enough then the cells undergo dielectric breakdown opening micropores in the cell membranes. dielectric breakdown seperates the charges of a dipole momentarily so that ions fill in the space in between the plates of the capacitor. The capacitor now discharges through the cells and the cells act as resistors to the current flow. During discharge there is likely problem that cell lysis can occur. The rate of lysing can be expressed as:
L= 1-(t/tc)exp(-(E-Ec)/K)
where tc= treshold value of treatment time
t= actual treatment time
Ec= treshold Efield strength
E= actual Efield strength
K= constant
QUESTION: CELL LYSIS IS AFFECTED BY WHICH OF THE FOLLOWING FACTORS
I. AN INCREASE IN CAPACITOR PLATE AREA
II. A DECREASE IN PLATE SEPERATION
III. INCREASING THE VALUE OF R1
IV. THE NUMBER OF CELLS BEING TREATED
ANSWER is all I,II,IV
I picked all of them and don't understand why R1 doesn't affect discharge. TBR explanation is charging R1 would affect only how long it takes the capacitor to charge up. it has no effect on discharging across the capacitor and the passage states that discharging the capacitor is what sends a current through the cells that may cause them to lyse