The circuit elements we have considered thus far are the emf (power supply or battery), the resistor (any device that converts electrical energy into thermal energy), and the capacitor (a device for storing separated charges). In circuit diagrams they are represented by the symbols shown on the right.

Most interesting circuits consist of more than just one
emf and one resistor. Some circuits have more than one resistor
connected in
**series**, as shown in the figure on the right. In such a circuit the
same current I flows through each resistor. The
potential difference across resistor 1 is V_{1 }= IR_{1}, the
potential difference across resistor 2 is V_{2 }= IR_{2}, and
the potential difference across resistor 3 is V_{3 }= IR_{3}. The potential difference across the chain is equal to the battery voltage V.
We have V = V_{1 }+ V_{2 }+ V_{3 }= IR_{1 }+ IR_{2
}+ IR_{3}. The **equivalent resistance**
of the circuit can be calculated from V = IR. The two equation yield

R = R_{1 }+ R_{2 }+ R_{3}.

The equivalent resistance of any number of resistors
connected in series is the sum of their individual resistances.

It is always
greater than any of the individual resistances.

In the figure above, let R_{1 }= 4 Ω, R_{2 }= 4 Ω, and R_{3
}= 12 Ω. Then the equivalent resistance is R = 20 Ω.

If the
battery voltage is 10 V, then the current in the circuit is I = V/R =
(10 V)/(20 Ω) = 0.5 A.

The current through each resistor is 0.5 A. The power
dissipated is P = IV = 5 W.

Assume another 20 Ω resistor is added to the chain. Adding another
20 Ω resistor increases the equivalent resistance to R = 40 Ω. The
current drops to I = (10 V)/(40 Ω) = 0.25 A. The power dissipated drops to
P = 2.5 W.

Adding an additional resistance in series

- reduces the current in the circuit,
- reduces the power I
^{2}R_{i}dissipated by each of the i resistances R_{i}, - reduces the total power dissipated.

The light bulbs in a chain of Christmas lights are often connected in series. If 10 identical light bulbs are in the chain and the voltage across the chain is 120V, then the voltage across each bulb is 12 V. The same current flows through each bulb. If one bulb burns out, no current flows, and none of the bulbs will light up.

It is more common to have more than one resistance connected in **
parallel**, as shown in the figure to the
right. The current I from the battery is divided into I_{1}, I_{2},
and I_{3} flowing through R_{1}, R_{2}, and R_{3},
respectively. These three currents recombine when the branches meet again.

I = I_{1 }+ I_{2 }+ I_{3}.

The voltage across each resistor in the parallel circuit shown is equal to
the battery voltage V.

We have I_{1 }= V/R_{1}, I_{2
}= V/R_{2}, I_{3 }= V/R_{3}. We find the
equivalent resistance from V = IR,

1/R = I/V = (I_{1 }+ I_{2 }+ I_{3})/V
= (1/R_{1}) + (1/R_{2}) + (1/R_{3}).

1/R = (1/R_{1}) + (1/R_{2}) +
(1/R_{3}).

For a set of parallel resistors the reciprocal of their
equivalent resistance equals the sum of the reciprocals of their individual
resistances.

Note: The equivalent resistance of any parallel combination of resistors is
always less than any of the individual resistances.

Assume that in the circuit shown on the right R_{1 }= 8 Ω, R_{2
}= 8 Ω, and R_{3 }= 4 Ω

Then 1/R = 1/(8 Ω) + 1/(8 Ω) + 1/(4 Ω) =
1/(2 Ω). R = 2 Ω.

If the battery voltage is 10 V, then the current in
the circuit is I = V/R = (10 V)/(2 Ω) = 5 A.

The individual currents are found
from I_{i }= V/R_{i}. The voltage across each resistor is 10
V, therefore

I_{1 }= (10 V)/(8 Ω) = 1.25 A,

I_{2 }=
(10 V)/(8 Ω) = 1.25 A ,

I_{3 }= (10 V)/(4 Ω) = 2.5 A .

The currents add to yield the total current I = 5 A.

If another 8 Ω resistor is added in parallel, a current of 1.25 A will
flow through this resistor. The total current in the circuit
increases, the equivalent resistance decreases, and the power P = IV
dissipated by the circuit increases.

If 10 Christmas lights are connected in parallel, then one burned-out bulb does not prevent the other bulbs from lighting up. However, a different type of bulb must be used if the bulbs are to be connected parallel to a 120 V outlet, because the voltage across each bulb will be 120 V.

Often circuits contain one emf and a number of series and parallel connections. An example is the section of a circuit shown on the right. The total resistance of such a circuit can be found by sequentially replacing parts of the circuit with a single equivalent resistance. The equivalent circuit then consists of a single emf and a single resistor. This procedure yields the total current flowing in the circuit, and the currents flowing through the individual resistors can be found by reversing the reduction process.

Let us find the total resistance between the points A and B in the
circuit to the right.

R_{1} and R_{2} are in series and are
replaced by R_{12 }= R_{1 }+ R_{2}.

Now R_{12} and R_{3} are in parallel and can be replaced
by R, where 1/R = 1/R_{12}+ 1/R_{3}.

If R_{1 }= R_{2
}= R_{3 }= 1 Ω, then R_{12 }=
2 Ω and R = (2/3) Ω.

If a 6 V battery is connected across points A and B, then
the current flowing in the circuit is I = V/R = 9 A.

The current flowing
through R_{3} is I = V/R_{3}, the current flowing through R_{1}
and R_{2} is V/R_{12 }= 3 A.

- Two (or more) resistors with their heads directly connected together and
their tails directly connected together are in parallel. They can be
replaced by one equivalent resistor R using 1/R = (1/R
_{1}) + (1/R_{2}) + (1/R_{3}) + .... - Two (or more) resistors connected together so that the tail of one is
connected to the head of the next, with no other path for the current to
take along the line connecting them, are in series. They can be replaced by
one equivalent resistor R using R = R
_{1 }+ R_{2 }+ R_{3}+.... - For resistors in series, the same current flows through each resistor, and for resistors in parallel, the same voltage drops across each resistor.

What are the largest and smallest resistances you can obtain by connecting a 36 Ω, a 50 Ω, and a 700 Ω resistor together?

Solution:

To obtain the largest resistance, we have to connect the resistors in
series.

R_{largest} = (36 + 50 + 700) Ω = 786 Ω.

To obtain the smallest resistance, we have to connect the resistors in
parallel.

1/R_{smallest} = (1/36 + 1/50 + 1/700) Ω^{-1} = 0.0492/Ω.

R_{smallest} = 20.32 Ω.

A** junction** in a circuit is a
point where at least three circuit paths meet. A
**branch** is a path connecting two junctions. The
circuit on the right has two junctions, labeled A and B, and three branches, namely the
three different paths from A to B. This circuit is a **multi-loop circuit**,
with more than one battery in different branches of the circuit. To
analyze such a circuit and to find the currents in all branches of the
multi-loop circuit one must use **Kirchhoff's rules**.

- Kirchhoff's first rule:

**Junction rule:**

At any junction point in a circuit where the current can divide, the sum of the currents into the junction must equal the sum of the currents out of the junction. (This is a consequence of charge conservation.) - Kirchhoff's second rule:

**Loop rule:**When any closed circuit loop is traversed, the algebraic sum of the changes in the potential must equal zero. (This is a consequence of conservation of energy.)

Kirchhoff's rules apply to all circuit.** **

- Replace any combination of resistors in series or parallel with their equivalent resistance. Label each remaining resistor and each emf with a symbol.
- Choose a direction for the current in each branch of the circuit, and label each current.
- Add plus and minus signs to label the high and low potential sides of
each circuit element, (i.e. each emf, resistor, capacitor). Label the
side of the resistor on which the current enters with a plus sign (+) and
the side on which the current exits with a minus sign (-).
- Sometimes it is hard to predict in which direction the current will flow in a particular loop. Simply pick a direction. If the real current flows in the opposite direction, the current in your solution will be negative. The minus sign just indicates that the current flows in the direction opposite to the one you picked.

- Apply Kirchhoff's first rule to all but one of the junctions in the
circuit.
- Each time you use the junction rule a current that has not been used before must be included.

- Apply Kirchhoff's second rule to as many loops as needed to obtain as
many equations (including the junction equations) as you have unknowns.
- Each time you use the loop equation you have to include a circuit element that has not been used before.
- To write down a loop equation, you choose a starting point, and then follow a path around the loop in one direction until you get back to the starting point. As you cross a battery or a resistors note the change in voltage. If you cross from - to +, the change is positive. If you cross from + to -, the change is negative. Add these changes in voltage and set the sum equal to zero.

- Solve the equations to obtain the values of the unknowns.

In the circuit
to the right, let V_{1} be a 1 V battery, V_{2} be a
2 V battery, V_{3} be a 3 V battery, and let each resistor be a 1 Ω resistor.
Pick directions for the currents as shown in the diagram.

- Junction A: I
_{1}+I_{3 }= I_{2} - Loop 1: V
_{1 }- I_{1}R_{1 }- I_{2}R_{2 }- V_{2 }- I_{1}R_{4 }= 0

This yields

1V - I_{1}1 Ω - I_{2}1 Ω - 2 V - I_{1}1 Ω = 0, or 2I_{1 }+ I_{2 }= -1 A. (Units: V/Ω = A) - Loop 2: V
_{3 }- I_{3}R_{3 }- I_{2}R_{2 }- V_{2 }= 0

This yields

3V - I_{3}1 Ω - I_{2}1 Ω - 2 V = 0, or I_{3 }+ I_{2 }= 1 A. - We now have to solve the three equations,

I_{1 }+ I_{3 }= I_{2},

2I_{1 }+ I_{2 }= -1 A,

I_{3 }+ I_{2 }= 1 A,

for the three unknown currents.- Eliminate I
_{3}: I_{3 }= 1 A - I_{2}using the third equation. The first two equations the yield

I_{1 }+ 1 A - I_{2 }= I_{2}, I_{ 1}- 2I_{2 }= -1 A,

2I_{1 }+ I_{2 }= -1 A. - Eliminate I
_{2}: I_{2 }= -1 A - 2I_{1}. We now can solve for I_{1}.

I_{1 }+ 2 A + 4I_{1 }= -1 A, 5I_{1 }= -3 A, I_{1 }= -(3/5) A. - Now we can solve for I
_{2}and I_{3}.

I_{2 }= -1 A - 2I_{1 }= -1 A + (6/5) A = (1/5) A

I_{3 }= I_{2 }- I_{1 }= (4/5) A

- Eliminate I
- I
_{2}and I_{3}flow in the directions picked in the diagram, I_{1}flows in a direction opposite to the one picked in the diagram.