Curriculum · Course · Week 4 · فارسی
KCL accounts for charge at a node: incoming and outgoing currents balance. KVL accounts for signed voltage changes around a closed loop. Together they turn a drawing into equations. A voltage divider is a convenient consequence for series resistors. Identify the actual nodes before applying a shortcut. An attached load changes the lower branch.
Choose reference current arrows and label one ground node. Keep those choices while solving. For a resistor, moving with its current gives a voltage drop of IR. For an unloaded divider, Vout=Vs R2/(R1+R2). A negative solved current means the real direction opposes your chosen arrow.
Use a 12 V source with 2 kΩ above 4 kΩ.
I=12/6k=2 mA, drops 4 V and 8 V, so Vout=8 V. Check 12−4−8=0.
A node receives 5 mA. Two branches carry 2 mA and an unknown outgoing current.
KCL gives 5=2+I, so I=3 mA. Every arrow is counted exactly once.
Reverse the direction used to traverse the first loop.
The equation becomes −12+8+4=0. Signs change together and the physical answer is unchanged.
Use ground at the bottom. Follow the source rise of 12 V, then the resistor drops of 4 V and 8 V. The output node is 8 V above ground. Reversing the loop traversal reverses all three signs together.
Keep the 12 V source and upper 2 kΩ resistor fixed. At R2=2 kΩ the output is 6 V. At 4 kΩ it is 8 V. Doubling R2 does not double the output because R2 also appears in the denominator.
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.
for lower_kohm in (1, 2, 4, 8):
current_ma = 12 / (2 + lower_kohm)
output_v = current_ma * lower_kohm
upper_drop_v = current_ma * 2
print(lower_kohm, "kohm:", round(output_v, 3), "V")
assert abs(12 - upper_drop_v - output_v) < 1e-10
Expected check: with R2=4 kΩ, Vout is 8 V and the KVL check is zero.
Kirchhoff’s current law follows from charge balance. In the steady lumped DC model, charge does not accumulate indefinitely at a node, so the algebraic sum of branch currents is zero. Choose all currents entering as positive or all leaving as positive and keep that convention. A negative solved current reverses your reference arrow; it does not invalidate the calculation.
Kirchhoff’s voltage law says the signed potential changes around a closed lumped-circuit path sum to zero. Starting from one potential and returning to the same point must give the same potential. Across a resistor, moving with its reference current gives a drop −IR when the current enters its positive-labeled terminal. Crossing an ideal source from minus to plus gives a rise +Vs. These rules require consistent reference choices, not a particular clockwise direction.
For unloaded series resistors, I=Vs/(R1+R2) and Vout=IR2, giving Vout=Vs R2/(R1+R2). The ratio is dimensionless. For positive resistances, output lies strictly between zero and Vs when Vs is positive. With a load RL across R2, replace R2 by Rp=R2 RL/(R2+RL). Current through R1 now divides between R2 and RL, so assuming the same current in all three components would be wrong.
A 12 V source drives R1=2 kΩ and R2=4 kΩ. A 4 kΩ load connects across R2. The unloaded output is 8 V. With the load, Rp=4 || 4=2 kΩ, so the output becomes 6 V. The upper current is (12−6)/2=3 mA. The lower resistor and load each carry 6/4=1.5 mA, satisfying KCL. Powers are 18 mW in R1 and 9 mW in each lower branch. Their sum, 36 mW, equals 12 V × 3 mA. Loading reduced output by 25% relative to 8 V.
A node V connects to a known 12 V node through 2 kΩ, a known 6 V node through 3 kΩ, and ground through 6 kΩ. Choose currents leaving V. KCL gives (V−12)/2+(V−6)/3+V/6=0, with volts and kilohms yielding milliamps. Multiply by 6 to get 3(V−12)+2(V−6)+V=0. Thus 6V−48=0 and V=8 V. Branch currents leaving are −2 mA, 0.667 mA and 1.333 mA. Their sum is zero. The negative first result means 2 mA actually enters from the 12 V node. This is introductory node-voltage reasoning using only KCL and Ohm’s law.
An ideal 12 V source and an ideal 5 V source oppose each other in a loop with 1 kΩ and 2 kΩ series resistors. Choose the loop current in the direction of the 12 V rise. Then 12−5−I(1+2)=0, giving I=7/3≈2.333 mA. The resistor drops are 2.333 V and 4.667 V. Reversing the reference direction gives −2.333 mA and the same physical flow. State source polarities explicitly; saying only “two sources in series” is insufficient.
Problems 1–6 are core. Problems 7–8 are optional design and reasoning extensions.
Common mistakes: changing reference directions halfway through a solution, treating crossing wires as a junction without checking their connection, and applying an unloaded divider formula to a loaded network. Finish with KCL, KVL and power checks wherever possible.
An original summary aligned with these textbook topics. References are optional further reading.