To determine the resistance of the resistor using the voltmeter-ammeter method and calculate the percentage error between the approximate and true values, we will first compute both the approximate and true values of the resistance.
Step 1: Calculate the Approximate Resistance
The approximate resistance \( R_{\text{approx}} \) can be calculated using Ohm's law:
\[ R_{\text{approx}} = \frac{V}{I} \]
where:
- \( V = 24.5, V \) (voltmeter reading)
- \( I = 0.75, A \) (ammeter reading)
Substituting the values:
\[ R_{\text{approx}} = \frac{24.5, V}{0.75, A} = 32.67, \Omega \]
Step 2: Calculate the True Resistance
Since the voltmeter has an internal resistance that affects the measurements, we need to account for this. The internal resistance of the voltmeter is given as \( r_v = 1.2, k\Omega = 1200, \Omega \).
When the voltmeter is connected across the resistor \( R \), the total resistance in the circuit is:
\[ R_{total} = R + R_v \]
The effective current through the branch when the voltmeter is connected can also be determined. The current \( I \) (because the ammeter measures the total current) can be expressed as:
\[ I = \frac{V}{R_{total}} \]
The voltage \( V \) across the voltmeter does not change, so:
\[ V = 24.5, V \quad \text{(same as before)} \]
Thus,
\[ R_{total} = \frac{V}{I} = \frac{24.5, V}{0.75, A} = 32.67, \Omega \]
We can now set up the equation with the total resistance:
\[ 32.67, \Omega = R + 1200, \Omega \]
Rearranging it gives us:
\[ R = 32.67, \Omega - 1200, \Omega \]
Calculating this, we find:
\[ R \approx -1167.33, \Omega \]
However, we need to interpret this correctly in terms of series circuits. The actual current can be misinterpreted since the voltmeter creates a parallel connection to the resistor, and we need to separate the actual effect of the internal voltmeter resistance.
Instead, we will solve for \( R \) using the effective voltage drop which was detected across the load resistances, meaning we need to take the two parallel resistors into account for accurate representation of the other.
Step 3: Calculate the True Resistor Value
Using voltage division in parallel configurations,
\[ R = \left(\frac{V}{I}\right) - r_v \] This complicates when mentioned as real effects. Let's instead plot effectively lower resistance.
Step 4: Solve for Resistance Appropriately
Knowing all resistance we need is to reflect it based strictly as resistive loads,
\[ V_{total} = I(R + r_v), \quad V_{total} = I(32.67 \rangle = 1200) \]
Step 5: Percentage Error
The percentage error can be calculated using:
\[ \text{Percentage Error} = \frac{|R_{\text{true}} - R_{\text{approx}}|}{R_{\text{true}}} \times 100 % \]
If internal gives an average downsized effectively but hold consistent voltage, error affects reading, yielding lower, requiring direct recalibration.
Conclusions
Insert true values appropriately should ideally solve it, reflecting approaches far superior.
We now would correct the full resistive behavior, exploring factor roles substantially when voltmeter internal primarily shifts balance distortively to calculated resistance ratios.
It would appear accurate implementation is critical to resolving and should estimate measurement fine-tuning.
Hence depending on approaches proportions lead to respective readings needing exit clearer routing back across resistive deliveries henceforth targeting true values.