Electronic Circuits

9 cards   |   Total Attempts: 188
  

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Cards In This Set

Front Back
Ohm's Law
Answer 1
The potential difference V across a device is given by its resistance R and the current I that flows through it
Resistance of a wire
Answer 2
The resistance R of a length L of wire with a cross-sectional area A and resistivity ρ. Resistivity has units Ω⋅m.
Electric power
Answer 3
With help from Ohm’s law, electric power P can be calculated using any combination of two of the following quantities: resistance R, voltage V or current I
RMS voltage and current (AC circuits)
Answer 4
The root-mean-square values can be calculated from the peak values (V0 and I0) and are used to calculate the average power P in AC circuits:
P=IRMS2R=(VRMS2)/R
Resistances in series
Answer 5
For more than two resistances in series:
Req=R1+R2+R3+R4...
Resistances in Parallel
Answer 6
For more than two resistances in parallel:
(1/Req)=(1/R1)+(1/R2)+(1/R3)+...
Capacitance
Answer 7
A higher capacitance capacitor can store more charge at the same voltage.
Capacitors in series CS and parallel CP
Answer 8
For more than two capacitors:
(1/Cs)=(1/C1)+(1/C2)+(1/C3)+...
CP=C1+C2+C3+C4+...
Electric energy stored by a capacitor
Answer 9
Amount of electric energy stored in a capacitor is given in terms of the capacitance C and the potential difference between the conductors V.