Chapter 6: Circuits

Question 1

Current (I)

Answer 1

I= Q/change in T

The magnitude of the current I is the amount of charge Q passing
through the conductor per unit time Δt, and it can be calculated as

I = Q/ change in T

SI Unit is Ampere.

Question 2

Kirchhoffs laws

Answer 2

I into junction = I leaving junction

At any point or junction in a circuit, the sum of currents directed into that point equals the sum of currents directed away from that point. This is an expression of conservation of electrical charge.

Loop rules

V source= Vdrop

Question 3

Resistors

Answer 3

R= ρ L/A

where ρ is the resistivity, L is the length of the resistor, and A is its cross-sectional area.

Question 4

Ohm’s Law

Answer 4

V=IR

where V is the voltage drop, I is the current, and R is the magnitude of the resistance,
measured in ohms (Ω). Ohm’s law is the basic law of electricity because it states that for a given magnitude of resistance, the voltage drop across the resistor will be proportional to the magnitude of the current.

Question 5

Power

Answer 5

P=W/t = change in E/t

The rate at which energy is dissipated by a resistor is the power of the resistor and can be calculated from

P=IV = I^2R= V2/R

where I is the current through the resistor, V is the voltage drop across the resistor, and R is the resistance of the resistor. Note that these different versions of the power equation can be interconverted by substitution using Ohm’s law (V = IR).

Question 6

Resistors in series

Answer 6

As the electrons flow through each resistor, energy is dissipated, and there is a voltage drop associated with each resistor. The voltage drops are additive; that is, for a series of resistors,

R1, R2, R3, ⋯ Rn in OHMS , the total voltage drop will be Vs = V1 + V2 + V3 + ⋯ + Vn

Because V = IR, we can also see that the resistances of resistors in series are also additive, such that Rs = R1 + R2 + R3 + ⋯ + Rn

Question 7

Resistors in parallel

Answer 7

Vp = V1 = V2 = V3 = ⋯ = Vn

Remember Kirchhoff’s loop rule: if every resistor is in parallel, then the voltage drop
across each pathway alone must be equal to the voltage of the source.

1/R= 1/R1 + 1/R2 + 1/R3

Question 8

Capacitors

Answer 8

Capacitors are characterized by their ability to hold charge at a particular voltage.
There are excellent real-world examples of capacitors. Perhaps the most important capacitor you’ll encounter in the clinics is the defibrillator.

If a voltage V is applied across the plates of a capacitor and a charge Q collects on it (with +Q on the positive plate and –Q on the negative plate), then the capacitance is given by: farad (g=9.8 m/s^2)

C=Q/V

The capacitance of a parallel plate capacitor is dependent upon the geometry of the two
conduction surfaces. For the simple case of the parallel plate capacitor, the capacitance is given by

C= ε0 (A/D)

is the permittivity of free space (8.85 x 10^-12)

Uniform electric field E= V/D

U=1/2 CV^2

Question 9

Dielectric materials

Answer 9

The capacitance due to a dielectric material is

C′ = κC

Incorporating the dielectric constant into Equation 6.14 reveals that capacitors are
CAκεd with charge (C = Aκε0/d).

A dielectric material can never decrease the capacitance; thus, κ can never be less
than 1.

Question 10

Capacitor in parallel

Answer 10

Therefore, Cp increases as more capacitors are added:

Cp = C1 + C2 + C3 + ⋯ + Cn

Just as we saw with resistors in parallel, the voltage across each parallel capacitor is the same and is equal to the voltage across the source.