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Examples: Electricity<^< Energy and Power | Course Index | Electronic Principles >^> Example 1:
A current of 1.5A is drawn from a 6V battery. What power is supplied?
Here we must use P = I x V (where I = 1.5A and V = 6V): P = I x V = 1.5A x 6V = 9W Hence a power of 9W is supplied. Example 2:
A voltage drop of 4V appears across a resistor of 200Ω. What power is dissipated in the resistor?
Here we use P = V2/R (where V = 4V and R = 200Ω): P = V2/R = (4V x 4V)/200Ω = 16/200 = 0.08W Hence the resistor dissipates a power of 0.08W (or 80mW). Example 3:
A current of 20mA flows in a 1kΩ resistor. What power is dissipated in the resistor?
Here we use P = I2 x R but, to note that we should convert mA and kΩ into A and Ω: P = I2 x R = (20mA x 20mA) x 1kΩ = (20 x 10-3 x 20 x 10-3) x 1 x 103 = 0.4W or 400mW. Thus a power of 400mW is dissipated by the resistor. <^< Energy and Power | Course index | Electronic Principles >^> |