Circuit

For the above circuit, I was asked to work out the voltage across resistor R2. The mark scheme calculated the total impedance (Z1+Z2), then done V/(Z1+Z2) = current; this value of current was then multiplied by the impedance of R2 and L combined (i.e. Z2). So the current multiplied by Z2 is equal to the voltage drop across R2?

I.E. voltage drop across Z2 (the combined) is equal to the voltage drop across R2?

Yes.

Because Z2 represents the parallel combination of R2 and L, the current flowing through Z2 is therefore the same as the sum of the currents in R2 and L.

And because R2 is in parallel with L, the pd across both is the same.

Ah thanks. One other question:



The above actually equals 2x10^-3 - (2.2 x10^-3) A

So shouldn't their answer be in mA rather than A?