but its usefulness will grow if you stay with electronics stuff:
Put simply, Kirchoff's Voltage Law states that the sum of the voltages around a closed loop will always equal zero.
So first - some rules:
- Assign a current, I, and stick with the direction
- Assign polarities to the components in your circuit.
- Voltage sources are "-" where the current goes in, and "+" where the current leaves.
- Voltage consumers (resistors) are "+" where the current goes in, and "-" where the current leaves.
- Write your equation, starting anywhere, and using the sign of the component where the current enters it.
So in the example above:
Since we know from Ohm's Law that Voltage = Current x Resistance:
So let's take a look at the result and see if it works:
So let's take a look at the result and see if it works:
So I = 1 amp
That makes the voltage drop across R1 = 1 amp x 8 Ω = 8 volts
Similarly, the voltage drop across R2 = 4 volts
So now we revisit the Kirchoff Voltage Loop and see that:
-12volts + 8volts +4volts = 0
One way this is useful is to consider the case of a blown fuse.
With the fuse blown, no current flows, but with the meter leads attached, a very small amount of current flows - indicated above by the grey arrows.
And we can use the Kirchoff Voltage Law to predict what the meter will read if the fuse is blown:
If Vm is the meter voltage, then the new loop becomes:
Now we find the current, I, in the circuit:
So the voltage drop across R1 = 12Ω x 1.2 micro amps (uA) = 14.4 micro volts (uV)
and the voltage drop across R2 = 8Ω x 1.2 uA = 9.6 uV
So now, without using the current to calculate the voltage across the meter resistance,
We can see that 12 volts - 14.4 micro volts - 9.6 micro volts - Vm = 0
And so Vm = 12 volts - 24 micro volts, which = 11.999976 volts.
Since most meters won't display 8 digits, the meter will read 12 volts.
Conclusion: If a fuse is blown in a series circuit, the voltage across the fuse will equal the source voltage.
Cheers for now.
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