Question

calculating resistance
at the test station you will measure the resistance of the red resistor and blue resistor shown in the circuits below. you will then create a circuit with both resistors and measure which resistor results in a larger voltage drop.

1. what will the resistance of red resistor be?
2. what will the resistance of blue resistor be?
3. what will the total resistance for the combined circuit be?
4. which resistor will cause a greater voltage drop in the combined circuit?

calculating voltage
at the test station you will measure voltage of the battery in the circuit below.

1. what will the voltage of the battery be?
2. was the voltage of the battery what you calculated? if not, why do you think the voltage might have been lower than expected?

Explanation:

Step1: Calculate red resistor resistance

Use Ohm's law $R=\frac{V}{I}$. For red resistor, $V = 1.5V$ and $I=0.075A$. So $R_{red}=\frac{1.5}{0.075}=20\Omega$.

Step2: Calculate blue resistor resistance

Using Ohm's law again. For blue resistor, $V = 1.5V$ and $I = 0.15A$. So $R_{blue}=\frac{1.5}{0.15}=10\Omega$.

Step3: Calculate total resistance of combined circuit

Since they are in series, $R_{total}=R_{red}+R_{blue}=20 + 10=30\Omega$.

Step4: Determine resistor with greater voltage - drop

In a series circuit, voltage - drop is proportional to resistance. Since $R_{red}>R_{blue}$, the red resistor will cause a greater voltage - drop.

Step5: Calculate battery voltage in second circuit

The two resistors are in series, total resistance $R = 5+5 = 10\Omega$, current $I = 0.3A$. Using Ohm's law $V=IR$, so $V=0.3\times10 = 3V$.

Answer:

1. $20\Omega$
2. $10\Omega$
3. $30\Omega$
4. Red resistor
5. $3V$
6. Without actual measurement data, we can't answer if it's the calculated value. Reasons for lower - than - expected voltage could include internal resistance of the battery, measurement errors, or non - ideal behavior of components.