Curriculum · Course · Week 3 · فارسی
A correct circuit equation can still give a wrong numerical answer if prefixes are mixed. Convert values into a consistent unit system before calculating. A voltmeter observes the difference between two nodes and is connected in parallel. An ideal voltmeter draws no current. A real meter adds a finite resistance and can change the circuit it measures. All experiments here are calculations or simulations.
Remember k = 10³, M = 10⁶, m = 10⁻³, and μ = 10⁻⁶. Uppercase M and lowercase m differ by a billion. In volts and kilohms, current comes out in milliamps. To model a voltmeter, add its input resistance between the measured nodes, then solve the new circuit.
Find the current for 5 V across 2.5 kΩ.
I=5/2500=0.002 A=2 mA. Its power is 10 mW.
A 10 V divider has two 10 kΩ resistors. Add a hypothetical 10 kΩ meter across the lower resistor.
The lower equivalent becomes 10k || 10k = 5k. The reading is 10×5/(10+5)=3.333 V, instead of 5 V. The meter loads the divider.
Repeat with a 10 MΩ input resistance.
10k || 10M≈9.990 kΩ and V≈4.9975 V. A larger input resistance reduces loading in this example.
The meter connects between Vout and ground. Model its input resistance as Rm in parallel with the lower 10 kΩ resistor. First solve the unloaded divider, then add Rm and solve again.
The horizontal axis is logarithmic. Moving one tick multiplies the meter resistance by ten. A 10 kΩ meter reads about 3.333 V. A 10 MΩ meter reads about 4.9975 V. Compare both to the ideal 5 V line.
For each problem, write the domain or units before calculating. After checking the answer, change one numerical value and solve the variant without looking back.
Open the full guided lab, graphs, circuit file and checkpoint
Predict the output by hand, then run the standard-library Python code below. Use your own calculation to explain any disagreement. Record at least three cases in your planner note.
import math
print("Half-value time:", 2 * math.log(2), "s")
for meter_kohm in (1, 10, 100, 1000, 10000):
lower_kohm = 10 * meter_kohm / (10 + meter_kohm)
output_v = 10 * lower_kohm / (10 + lower_kohm)
print(meter_kohm, "kohm meter:", round(output_v, 4), "V")
Expected check: the half-value time is about 1.386 s and the 10 kΩ meter reads about 3.3333 V.
This week combines unit discipline with a model of the measuring instrument. A plausible-looking number can be wrong by a factor of a thousand. Write the numerical value, prefix and unit together throughout the calculation. In engineering notation the exponent is a multiple of three, so 0.000047 A=47 × 10⁻⁶ A=47 μA. Prefixes are case-sensitive: M means mega and m means milli.
Dimensional checking asks whether the units follow the same algebra as the numbers. Since Ω=V/A, the expression V/Ω has units A, while V×Ω cannot represent current. Dimensional agreement is necessary but not sufficient: both an incorrect factor of two and a sign error can survive a unit check.
Let R1 and R2 form a divider across Vs. A voltmeter with input resistance Rm connects across R2. First form Rp=R2 Rm/(R2+Rm), then use the series current I=Vs/(R1+Rp). The indicated voltage is Vm=I Rp. This is a preview of next week’s general divider formula derived directly from last week’s network rules. There is no need to assume the divider formula beforehand.
An ideal voltmeter has infinite input resistance and does not divert current. An ideal ammeter has zero resistance and is inserted in series. A real ammeter adds series resistance and changes the current it measures. An ohmmeter applies its own test stimulus; resistance measurements are modeled with the external circuit de-energized. All tasks here remain software-only.
For a 10 V source and two 10 kΩ resistors, the unloaded midpoint is 5 V. A 100 kΩ meter makes the lower equivalent 10×100/(10+100)=9.0909 kΩ. The current becomes 10/(10+9.0909)=0.52381 mA, and the reading is 4.76190 V. The signed loading error is (4.76190−5)/5 × 100%=−4.7619%. The meter can be perfectly calibrated and still cause this error by changing the circuit.
An ideal 5 V source drives 1 kΩ. Without the meter, current is 5 mA. A meter modeled as 20 Ω in series changes total resistance to 1020 Ω. The measured current is 5/1020≈4.902 mA, a decrease of about 1.96%. The voltage across the meter is 0.004902×20≈0.0980 V. That additional voltage drop is the burden of this simplified meter model.
Suppose a display reports 4.76 V with a stated accuracy of ±(0.5% of reading + 2 counts), and its least significant count is 0.01 V. The bound is 0.005×4.76+2×0.01=0.0438 V. Report approximately 4.76±0.044 V under those stated conditions. Extra display digits do not remove loading error. Accuracy specifications and loading describe different effects and should be recorded separately.
Problems 1–6 are core. Problems 7–8 are optional extensions.
Common mistakes: confusing resolution with accuracy, treating a meter as invisible, putting an ammeter in parallel, and mixing MΩ with mΩ. Always draw the meter’s equivalent resistance before recalculating.
An original summary aligned with these textbook topics. References are optional further reading.