02 // Visual Math & Finance SolversI = V / R, P = V × I = I² × R = V² / R
Visual Ohm's Law ($V = I \times R$) & DC Circuit Power Schematic Lab
Ohm's Law states that steady-state current $I$ through a conductor is directly proportional to potential difference $V$ and inversely proportional to resistance $R$ ($I = V/R$), while dissipated thermal power obeys Joule's Law $P = VI = I^2R = V^2/R$.
I = V / R, P = V × I = I² × R = V² / RInputs & Parameters (3)
V
1V±1240V
Ω
0.5Ω±0.5500Ω
h
0.5h±0.524h
Current I = V / R
2 A
2,000 mA steady-state DC flow
Power P = V × I = I²R
24 W
0.024 kW thermal dissipation
Daily Energy (8h/day)
0.192 kWh
Monthly (30d) = 5.76 kWh
Conductance G = 1 / R
0.1667 S
Load Resistance R = 6 Ω
20-Row Comparison Table
Compare across:
| # | Source Voltage (V) (V) | Thermal Power (P) | Current I (A) | Daily Energy |
|---|---|---|---|---|
| 01 | 1 V | 0.167 W | 0.167 A | 0.001 kWh |
| 02 | 14 V | 32.667 W | 2.333 A | 0.261 kWh |
| 03 | 26 V | 112.667 W | 4.333 A | 0.901 kWh |
| 04 | 39 V | 253.5 W | 6.5 A | 2.028 kWh |
| 05 | 51 V | 433.5 W | 8.5 A | 3.468 kWh |
| 06 | 64 V | 682.667 W | 10.667 A | 5.461 kWh |
| 07 | 76 V | 962.667 W | 12.667 A | 7.701 kWh |
| 08 | 89 V | 1,320.167 W | 14.833 A | 10.561 kWh |
| 09 | 102 V | 1,734 W | 17 A | 13.872 kWh |
| 10 | 114 V | 2,166 W | 19 A | 17.328 kWh |
| 11 | 127 V | 2,688.167 W | 21.167 A | 21.505 kWh |
| 12 | 139 V | 3,220.167 W | 23.167 A | 25.761 kWh |
| 13 | 152 V | 3,850.667 W | 25.333 A | 30.805 kWh |
| 14 | 165 V | 4,537.5 W | 27.5 A | 36.3 kWh |
| 15 | 177 V | 5,221.5 W | 29.5 A | 41.772 kWh |
| 16 | 190 V | 6,016.667 W | 31.667 A | 48.133 kWh |
| 17 | 202 V | 6,800.667 W | 33.667 A | 54.405 kWh |
| 18 | 215 V | 7,704.167 W | 35.833 A | 61.633 kWh |
| 19 | 227 V | 8,588.167 W | 37.833 A | 68.705 kWh |
| 20 | 240 V | 9,600 W | 40 A | 76.8 kWh |
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Key Analytical Takeaways
- 01Doubling voltage across a fixed resistor quadruples power dissipation ($P = V^2 / R$) because both $V$ and $I$ double simultaneously.
- 02The live circuit schematic animates current intensity and thermal load glow in proportion to $I = V/R$ and wattage $P$.
- 03All 12 algebraic wheel identities connecting $(V, I, R, P)$ are evaluated with live numerical values in the quadrant wheel.
Frequently Asked Questions
Why do high-voltage transmission lines reduce power loss?+
To transmit a fixed power P = V·I, stepping up voltage V reduces line current I proportionally. Since resistive wire loss scales with I²R, cutting current by 10× reduces heat loss in the cables by 100×.
How are Watts converted into kilowatt-hours (kWh)?+
Multiply power in Watts by operating hours and divide by 1,000: Energy (kWh) = (Watts × Hours) / 1,000.
Why does doubling voltage quadruple resistive power dissipation?+
By Ohm's Law (I = V / R), doubling voltage across a fixed resistance R simultaneously doubles the current I, so dissipated thermal power P = V × I = V² / R scales quadratically (4×).