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The specified current in the problem text is the total current `I_T =10 A` shown in the figure. Because there are 3 components of the current (total, resistive and reactive), the voltage drop over all the components of the circuit is the same. It means the circuit is parallel.
Now, from the figure posted with the problem text, one can see that the resistive component of the current is just
`I_R = I*T*cos(theta) = 10*cos(36.9) = 8 A`
where `theta =36.9 degree` is the phase difference between the total and resistive currents.
From the currents right triangle one can now determine the reactive component of the current
`I_X = sqrt(I_T^2-I_R^2) = sqrt(10^2 -8^2) = 6 A`
The resistive current is I(R) = 8 A and the reactive current is I(X) =6 A.
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