The circuit shown in the figure consists of resistance, capacitance and inductance in series with a 100V source when the
Posted: Thu Jul 14, 2022 10:59 am
a) 100 = 20i + 0.05 \(\frac{di}{dt} + \frac{1}{20 \times 10^{-6}} \int idt\)
b) 100 = 20i – 0.05 \(\frac{di}{dt} + \frac{1}{20 \times 10^{-6}} \int idt\)
c) 100 = 20i + 0.05 \(\frac{di}{dt} – \frac{1}{20 \times 10^{-6}} \int idt\)
d) 100 = 20i – 0.05 \(\frac{di}{dt} – \frac{1}{20 \times 10^{-6}} \int idt\)
b) 100 = 20i – 0.05 \(\frac{di}{dt} + \frac{1}{20 \times 10^{-6}} \int idt\)
c) 100 = 20i + 0.05 \(\frac{di}{dt} – \frac{1}{20 \times 10^{-6}} \int idt\)
d) 100 = 20i – 0.05 \(\frac{di}{dt} – \frac{1}{20 \times 10^{-6}} \int idt\)