# Measured voltage on coil

I have a simple circuit of coil with inductance L $$u_c(t)= -L\frac{di}{dt}$$ and AC source with output voltage $u_s$ What is actually measured by the voltmeter ($u_c$, $u_s$ or $u_s - u_c$)?

If you know that the induced voltage is opposite to the source voltage, how would you explain that the measured voltage is not equal 0.

• what do you think? include in the question what you have thought so far and what exactly confuses you – glS Jan 24 '15 at 15:55

According to Kirchhoff law's, all voltages in the loop must sum up to 0, thus $$u_c = u_s = L\frac{\mathrm{d}i}{\mathrm{d}t}$$
Interesting is what happens with current. After small math you get $$\int\mathrm{d}i = i(t) = \frac{1}{L}\int u_s \mathrm{d}t$$ If source is a sinusoidal of amplitude V, then $$i(t) = -\frac{V}{\omega L}\cos{\omega t}$$
• Isn't KVL law for this circuit $u_c + u_s = 0$, so $u_s = - u_c$ – Peter Jan 25 '15 at 9:12
• $u_s$ is an active voltage source, so it adds to the circuit, and $u_c$ is a passive element - it "receives" voltages, so is with minus. For example, RLC circuit with two sources you would write $$u_{s1} + u_{s2} - u_r - u_l - u_c = 0$$It is similar to vector adding, draw voltage drops vectors on each element and you will find it. – Rafal Jan 25 '15 at 10:22
• That depends on how you define $u_c$, right? Taking your answer $u_c = u_s = -L\frac{\mathrm{d}i}{\mathrm{d}t}$, we have $0 = u_s - u_c = u_s - (-L \frac{\mathrm{d}i}{\mathrm{d}t}) = u_s + L \frac{\mathrm{d}i}{\mathrm{d}t}$ – Peter Jan 25 '15 at 12:03
• I think you have it wrong, see electronics-tutorials.ws/inductor/ac-inductors.html $u_s \sim \sin{\omega t}$ and $i \sim -\cos{\omega t}$ – Peter Jan 25 '15 at 15:27