Day 117

Series Circuit Example (part 1)

Given: R₁ = 1Ω, R₂=2Ω, R₃=3Ω,

V (battery) = 12V

Find the current through each R and the V drop across each resistor. First, find the total resistance, which is the sum total of all three = 6Ω = R

Series Circuit Example (part 2)

Use Ohm’s Law to find the total current

I = V/R = 12V/6Ω = 2A

Since it is a series circuit, the current through each R is the same, so I1 = I2 = I3 = 2A

Series Circuit Example (part 3)

The voltage drop across each R is also found by Ohm’s Law

Note that 2V + 4V + 6V = 12V

Series Circuit Example (part 4)

To find the potential after the current goes through each R, start with a 12V potential at the positive terminal.

After the current goes through the 1Ω R, the voltage drops 2V and the new potential after the R is 10V.

After the 2Ω R, the voltage drops 4V, so potential is now 6V.

After the 3Ω R, the voltage drops 6V, so potential is now 0V and is now connected to the negative terminal. The cycle starts again.

Galvanometer

Is an instrument that is capable of measuring only very small currents.
Principle is a torque acting on a current loop in the presence of a magnetic field; amount of current is proportional to amount of torque
Can be adjusted so as to become an ammeter capable of reading higher currents.

Galvanometer Turned Ammeter [G A]

A Galvonometer should be wired in parallel with a “shunt resistor” The “shunt resistor” carries most of the current The entire two-branch system becomes the new ammeter.
Need to decrease the large resistance of a Galvanometer and need to change the small scale used To accomplish this, place the shunt resistor, Rrp, in parallel with the Galvanometer.

Voltmeter Example (part 1)

Given: A Galvanometer with a R=80Ω has a maximum deflection of 1 mA. The largest deflection that can be measured is V(max)=(0.001A)(80Ω)=0.08V Find the resistor that must be added to the Galvanometer to form a Voltmeter capable of reading 80V (this will be Rs)