A typical cell membrane is 8.0 nm thick and has an electrical resistivity of 1.3 × 107Ω.m.
(i) If the potential difference between the inner and outer surfaces of a cell membrane is 75 mV, how much current flows through a square area of membrane on a side?
(ii) Suppose the thickness of the membrane is doubled, but the resistivity and potential difference remain the same. Does the current increase or decrease? By what factor?
(i) Well, well, looks like our cell membrane decided to get all electric on us! So, we have a thickness of 8.0 nm and a potential difference of 75 mV. Let's calculate the current.
First, we need to convert the thickness from nanometers to meters. 1 nm is equal to 1 × 10^-9 m. So, our thickness is 8.0 × 10^-9 m.
Next, we use Ohm's Law, which states that current (I) equals potential difference (V) divided by resistance (R). In this case, our resistance is given by resistivity (ρ) multiplied by length (L) divided by area (A). Phew, that was a mouthful!
Let's plug in the values:
Resistance = resistivity × length / area = (1.3 × 10^7 Ω.m) × (8.0 × 10^-9 m) / A
We are looking for the current, so let's rearrange the formula:
Current = Potential difference / Resistance = 75 mV / [(1.3 × 10^7 Ω.m) × (8.0 × 10^-9 m) / A]
Now, we can simplify a bit:
Current = 75 mV × A / [(1.04 × 10^-1 Ω) × 10^-9 m]
And simplify a bit more:
Current = 7.2 × 10^10 A × A
So, the current flowing through the square area of the membrane on each side is 7.2 × 10^10 A times the area. Now that's a shockingly high current!
(ii) Ah, so the cell membrane decided to go on a thickness-changing spree! If the thickness is doubled but the resistivity and potential difference stay the same, let's see what happens to the current.
Doubling the thickness means our new thickness is 2 × 8.0 nm = 16 nm.
But don't worry, I'm not going to double your workload. We can simply adjust the area of the membrane instead. Since the thickness doubled, let's halve the area.
So, if the thickness is doubled, but the resistivity and potential difference remain the same, the current decreases by a factor of 2.
But hey, don't give up hope. Remember, thicker or thinner, a cell membrane will always keep current events interesting!
To calculate the current flowing through a cell membrane, you can use Ohm's Law (V = IR), where V is the potential difference, I is the current, and R is the resistance.
(i) Let's start by finding the resistance of the membrane. The resistance (R) can be determined using the formula R = ρL/A, where ρ is the resistivity, L is the thickness of the membrane, and A is the area of the membrane.
Given:
Thickness of the membrane (L) = 8.0 nm = 8.0 × 10^(-9) m
Resistivity (ρ) = 1.3 × 10^(7) Ω.m
Using the formula, we can calculate the resistance (R) of the membrane:
R = ρL/A
Since we want to find the current flowing through a square area of the membrane, we need to determine the area (A). The area of a square is given by A = side length squared.
Let's assume the length of each side of the square is s. Therefore, the area will be A = s^2.
Given:
Potential difference (V) = 75 mV = 75 × 10^(-3) V
Now we have all the required values to calculate the resistance and current.
Using the resistance formula, R = ρL/A, we can rearrange it to solve for A:
A = ρL/R
A = (1.3 × 10^(7) Ω.m) × (8.0 × 10^(-9) m) / (75 × 10^(-3) V)
A ≈ 1.3867 × 10^(-8) m^2
Now, we can calculate the current (I) using Ohm’s Law:
I = V/R
I = (75 × 10^(-3) V) / (1.3 × 10^(7) Ω.m × 8.0 × 10^(-9) m / (75 × 10^(-3) V))
I ≈ 4.8077 × 10^(-7) A
Therefore, the current flowing through the square area of the membrane on a side is approximately 4.8077 × 10^(-7) A.
(ii) If the thickness of the membrane is doubled but the resistivity and potential difference remain the same, we need to compare the current before and after the change.
Let's assume the initial current is I1, with a membrane thickness of L1. After doubling the thickness, the new current will be I2, with a membrane thickness of L2 = 2L1.
Using Ohm's Law, the initial current (I1) can be calculated as I1 = V/R1, and the new current (I2) can be calculated as I2 = V/R2.
Since the resistivity and potential difference remain the same, the resistance before (R1) and after (R2) the change can be calculated using R = ρL/A.
For R1, the initial resistance:
R1 = ρL1/A = ρL1/(s^2)
For R2, the resistance after the thickness is doubled:
R2 = ρL2/A = ρ(2L1)/(s^2)
To find the factor by which the current changes, we can compare I2 to I1.
Using Ohm's Law:
I1 = V/R1
I2 = V/R2
Dividing the two equations:
I2/I1 = (V/R2) / (V/R1)
I2/I1 = R1/R2
Substituting the values for R1 and R2:
I2/I1 = ρL1/A / (ρ(2L1)/A)
I2/I1 = L1 / (2L1)
I2/I1 = 1/2
Therefore, the current (I2) decreases by a factor of 2 when the thickness of the membrane is doubled, while the resistivity and potential difference remain the same.
To calculate the current flowing through a square area of the cell membrane, we can use Ohm's Law, which states that the current (I) through a conductor is equal to the potential difference (V) across it divided by its resistance (R):
I = V / R
(i) To find the current flowing through a square area of the membrane, we need to calculate the resistance of the membrane and use the given potential difference.
1. Start by calculating the resistance of the membrane. The resistance of the membrane (R) can be determined using the formula:
R = resistivity (ρ) * length (L) / area (A)
Given:
Length (L) = thickness of the membrane = 8.0 nm = 8.0 × 10^-9 m
Area (A) = area of the square membrane = side length^2
Potential difference (V) = 75 mV = 75 × 10^-3 V
Resistivity (ρ) = 1.3 × 10^7 Ω.m
2. Calculate the area of the square membrane (A). Since it is a square, the formula simplifies to:
A = side length^2
3. Substitute the given values to calculate the area:
A = (8.0 × 10^-9 m)^2
4. Calculate the resistance (R) using the resistivity, length, and area:
R = (1.3 × 10^7 Ω.m) * (8.0 × 10^-9 m) / [(8.0 × 10^-9 m)^2]
5. Calculate the current (I) using Ohm's Law:
I = (75 × 10^-3 V) / R
(ii) To determine the effect of doubling the membrane thickness on the current, we can compare the new resistance with the initial resistance.
1. Double the thickness of the membrane:
New thickness = 2 * 8.0 nm = 16.0 nm = 16.0 × 10^-9 m
2. Calculate the new resistance (R) using the same resistivity, new thickness, and same area:
New R = (1.3 × 10^7 Ω.m) * (16.0 × 10^-9 m) / [(8.0 × 10^-9 m)^2]
3. Compare the new resistance (New R) with the initial resistance (R) to determine if the current increases or decreases:
If New R > R, then the current decreases.
If New R < R, then the current increases.
4. Calculate the factor by which the current changes:
Change in current factor = R / New R
By following these steps, you can calculate the current through a square area of the cell membrane and determine how it changes when the membrane thickness doubles while keeping the resistivity and potential difference constant.
R = 1.3 * 10^7 * 8 * 10^-9 ohms/m^2
i = V/R
b. Same V, double R, half i