The quote you might be thinking of is: “The larger the cell is, the smaller the surface area to volume ratio.” This concept is often discussed in biology to explain why cells tend to remain small; as they grow larger, their volume increases at a faster rate than their surface area, which can impact their ability to transport nutrients and waste efficiently.
finish this quote, “ the larger the cell is, the smaller…”
3 answers
Consider the circuit below. What is the value of the current I? Write your answer in proper engineering units.
V1 5V
R1 1 k ohms
R2 2.2 k ohms
R3 1 k ohms
V1 5V
R1 1 k ohms
R2 2.2 k ohms
R3 1 k ohms
To find the current \( I \) flowing in the circuit you described, we first need to analyze the resistors and voltage source in the circuit.
Assuming the resistors \( R1 \), \( R2 \), and \( R3 \) are connected in series, the total resistance \( R_{total} \) can be calculated as follows:
\[ R_{total} = R1 + R2 + R3 \]
Substituting the values of the resistors:
\[ R_{total} = 1 , \text{k}\Omega + 2.2 , \text{k}\Omega + 1 , \text{k}\Omega \] \[ R_{total} = 1 + 2.2 + 1 = 4.2 , \text{k}\Omega \]
Next, we can use Ohm's law to find the current \( I \). According to Ohm's law:
\[ I = \frac{V}{R} \]
Where:
- \( V \) is the total voltage (5 V)
- \( R \) is the total resistance (4.2 kΩ)
Substituting the known values:
\[ I = \frac{5 , \text{V}}{4.2 , \text{k}\Omega} = \frac{5}{4200} \text{ A} \]
Calculating this gives:
\[ I = \frac{5}{4200} = 0.00119047619 \text{ A} \]
Converting this to milliAmperes (mA):
\[ I \approx 1.19 , \text{mA} \]
So the value of the current \( I \) is approximately:
\[ I \approx 1.19 , \text{mA} \]