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Examples: Electricity

<^< Energy and Power | Course Index | Electronic Principles >^>(:nl:)

>>frame round<< Example 1:

A current of 1.5A is drawn from a 6V battery.

What power is supplied? >><<

>>id=example1 class=question margin=10px frame round<< 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. >><<

>>frame round<< Example 2:

A voltage drop of 4V appears across a resistor of 200&#937;.

What power is dissipated in the resistor? >><<

>>id=example2 class=question margin=10px frame round<< Here we use P = V'^2^'/R (where V = 4V and R = 200&#937;):

P = V'^2^'/R = (4V x 4V)/200&#937; = 16/200 = 0.08W

Hence the resistor dissipates a power of 0.08W (or 80mW). >><<

>>frame round<< Example 3:

A current of 20mA flows in a 1k&#937; resistor.

What power is dissipated in the resistor? >><<

>>id=example3 class=question margin=10px frame round<< Here we use P = I'^2^' x R but, to note that we should convert mA and k&#937; into A and &#937;:

P = I'^2^' x R = (20mA x 20mA) x 1k&#937;

= (20 x 10'^-3^' x 20 x 10'^-3^') x 1 x 10'^3^'

= 0.4W or 400mW.

Thus a power of 400mW is dissipated by the resistor. >><<(:nl:)(:table style="clear:both":)