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An explanation of how to calculate currents and voltages in circuit components that are in series, parallel, or not simply in series or parallel using kirchoff’s rules. It includes examples of how to apply these rules to different circuits and how to find the equivalent resistance of resistors in series and parallel. The document also includes practice problems for the reader to try.
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Today’s agenda:
Resistors in Series and Parallel. You must be able to calculate currents and voltages in circuit components in series and in parallel.
Kirchoff’s Rules. You must be able to use Kirchoff’s Rules to calculate currents and voltages in circuit components that are not simply in series or in parallel.
There are ―two‖ ways to connect circuit elements.
Series:
A (^) B
Put your finger on the wire at A. If you can move along the wires to B without ever having a choice of which wire to follow, the circuit components are connected in series.
Truth in advertising: it is possible to have circuit elements that are connected neither in series nor in parallel. See problem 24.73 in the 12th^ edition of our text for an example with capacitors.
Are these resistors in series or parallel?
It matters where you put the source of emf.
+ -
V
parallel
Are these resistors in series or parallel?
It matters where you put the source of emf.
+
series
Here’s a circuit with three resistors and a battery:
+ -
V I
Current flows…
…in the steady state, the same current flows through all resistors…
…there is a potential difference (voltage drop) across each resistor.
Applying conservation of energy allows us to calculate the equivalent resistance of the series resistors.
I am including the derivation in these notes, for the benefit of students who want to look at it.
In lecture, I will skip ahead past the derivation.
Now imagine replacing the three resistors by a single resistor, having a resistance R such that it draws the same current as the three resistors in series.
+ -
I V
qV = qV 1 + qV 2 + qV 3
Req
+ -
I V
As above: V = IReq
From before: V = IR 1 + IR 2 + IR 3
Combining: IReq = IR 1 + IR 2 + IR 3
Req = R 1 + R 2 + R 3
For resistors in series, the total resistance is the sum of the separate resistances.
+ -
I^ V
Current flows…
…different currents flows through different resistors…
…but the voltage drop across each resistor is the same.
Here’s another circuit with three resistors and a battery.
Applying conservation of charge allows us to calculate the equivalent resistance of the parallel resistors.
I am including the derivation in these notes, for the benefit of students who want to look at it.
In lecture, I will skip ahead past the derivation.
Now imagine replacing the three resistors by a single resistor, having a resistance R such that it draws the same current as the three resistors in parallel.
Req
+ -
I^ V
A B
From above, I = I 1 + I 2 + I 3 , and
1 2 3 1 2 3
So that eq 1 2 3
Dividing both sides by V gives
eq 1 2 3
a consequence of eq i i conservation of charge
Note: for resistors in parallel, Req is always less than any of the Ri.
I’ll work this ―conceptually.
Here’s the key to solving Physics problems: don’t bite off more than you can chew. Bite off little bite-sized chunks.
Example: calculate the equivalent resistance of the resistor ―ladder‖ shown. All resistors have the same resistance R.
A hot dog. Where do you take the first bite?
