Question

In the circuit given below, the KVL for first loop is ___________

(a) V (t) = R1i1 + L1 \(\frac{di_1}{dt}\) + M \(\frac{di_2}{dt}\)

(b) V (t) = R1i1 – L1 \(\frac{di_1}{dt}\) – M \(\frac{di_2}{dt}\)

(c) V (t) = R1i1 + L1 \(\frac{di_1}{dt}\) – M \(\frac{di_2}{dt}\)

(d) V (t) = R1i1 – L1 \(\frac{di_1}{dt}\) + M \(\frac{di_2}{dt}\)

Answer 1

Right answer is (c) V (t) = R1i1 + L1 \(\frac{di_1}{dt}\) – M \(\frac{di_2}{dt}\)

To explain: We know that, in general, the KVL is of the form V (t) = R1i1 + L1 \(\frac{di_1}{dt}\) + M \(\frac{di_2}{dt}\)

But here, M term is negative because i1, is entering the dotted terminal and i2, is leaving the dotted terminal.

So, V (t) = R1i1 + L1 \(\frac{di_1}{dt}\) – M \(\frac{di_2}{dt}\).