Curriculum · Course · Week 2 · فارسی
V = IR; P = VI = I²R
Example 1. With 6 V across 2 kΩ, current is 3 mA. Use 0.003 A when multiplying by 6 V to obtain 18 mW.
Example 2. Two 2 kΩ resistors in series give 4 kΩ and carry the same current. In parallel they give 1 kΩ and have the same voltage. With 8 V, the source supplies 2 mA in series or 8 mA in parallel.
Practice. At 4 V, find the total current for the same pair in each arrangement. Answers: 1 mA in series, 4 mA in parallel.
Put 2 kΩ and 3 kΩ in parallel across 6 V. Both see 6 V. The branch currents are 3 mA and 2 mA, so the source supplies 5 mA. The equivalent resistance is 6 V / 0.005 A = 1.2 kΩ. It is below 2 kΩ, as expected for this parallel pair. The branch powers are 18 mW and 12 mW, which add to the source's 30 mW.
Answers. 1. 2 mA and 16 mW. 2. 3 mA, with drops of 3 V and 9 V. 3. 12 mA and 4 mA, 16 mA total, and 750 Ω.
The vertical axis is voltage and the horizontal axis is current. The steeper blue line is the 2 kΩ resistor. At 2 mA it needs 4 V, while the 1 kΩ resistor needs 2 V. At a fixed voltage, the higher resistance therefore carries less current. If you swap the axes, the slope becomes conductance.
Read the graph. To compare currents at the same voltage, trace a horizontal line from 2 V. It meets the 2 kΩ line at 1 mA and the 1 kΩ line at 2 mA. The ratio V/I gives resistance. For example, 2 V divided by 1 mA is 2 kΩ. Convert milliamps to amps if you want the result in ohms.
Ohm’s law V=IR models an ohmic resistor under specified physical conditions. It is not a universal statement that every component has constant resistance. A diode or heated filament can have a nonlinear voltage–current curve. In this week, assume positive, constant resistances and ideal connecting wires.
For series resistors the same current passes through every element. Adding the voltage drops gives Vs=IR1+IR2=I(R1+R2), so Req=R1+R2. In parallel every branch has the same voltage. Adding branch currents gives Is=V/R1+V/R2, hence 1/Req=1/R1+1/R2. For two resistors this becomes Req=R1R2/(R1+R2). That product-over-sum formula is not the general formula for three resistors.
Power measures energy transfer rate. Starting with P=VI, substituting V=IR gives P=I²R, and substituting I=V/R gives P=V²/R. Decide what is held constant before comparing power. Doubling R at fixed voltage halves power; doubling R at fixed current doubles it. Over time t at constant power, energy is E=Pt, in joules when power is watts and time is seconds.
A 1 kΩ resistor is in series with a parallel pair of 2 kΩ and 3 kΩ across an ideal 12 V source. First reduce the pair to 2×3/(2+3)=1.2 kΩ. Total resistance is 2.2 kΩ, giving Is=12/2.2=5.455 mA. The 1 kΩ drop is 5.455 V, leaving 6.545 V across each parallel resistor. Branch currents are 6.545/2=3.273 mA and 6.545/3=2.182 mA. Their sum recovers the source current.
The source delivers approximately 12×5.455=65.45 mW. Individual dissipations are 29.75 mW, 21.42 mW and 14.28 mW. Their sum is 65.45 mW to rounding. These are two independent checks: branch-current balance and energy balance. A result that passes only a calculator check is not yet a verified circuit solution.
An ideal 12 V supply feeds a 1 kΩ resistor with ±5% resistance tolerance. The resistance can lie between 950 Ω and 1050 Ω. Maximum current occurs at the smallest resistance: Imax=12/950≈12.63 mA. Maximum power is Pmax=12²/950≈0.1516 W. A 0.125 W nominal rating would be below this calculated worst case. A 0.25 W rating exceeds it, but a real design must also consider the manufacturer’s temperature derating and operating conditions. This is a paper design comparison, not a hardware construction task.
Solve core problems 1–6 and check units and plausibility. Problems 7–8 are optional extensions.
Common mistakes: adding parallel resistances, using milliamps as amps in a power product, and reducing a network from visual proximity rather than shared nodes. Keep a table of each component’s voltage, current and power.