Physics

If $$L$$ is the inductance, $$'i'$$ is current in the circuit, $$\dfrac { 1 }{ 2 } { Li }^{ 2 }$$ has the dimensions of:


ANSWER

Work


SOLUTION
$$L \rightarrow M{ L }^{ 2 }{ I }^{ -2 }{ T }^{ -2 }$$
$$ I\rightarrow I$$
$${ LI }^{ 2 } = (M{ L }^{ 2 }{ I }^{ -2 }{ T }^{ -2 }){ I }^{ 2 }= M{ L }^{ 2 }{ T }^{ -2 } \rightarrow  (unit \ of \ work)$$
$$Power  =  f\times v = ML{ T }^{ -2 } \times  L{ T }^{ -1 }\quad =\quad M{ L }^{ 2 }{ T }^{ -3 }$$
$$ Pressure = f/A = \dfrac { ML{ T }^{ -2 } }{ { L }^{ 2 } }  = { ML }^{ -1 }{ T }^{ -2 }$$
$$force = { M }^{ 1 }{ L }^{ 1 }{ T }^{ -2 }$$
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