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
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 V1 = IR1, the
potential difference across resistor 2 is V2 = IR2, and
the potential difference across resistor 3 is V3 = IR3. The potential difference across the chain is equal to the battery voltage V.
We have V = V1 + V2 + V3 = IR1 + IR2
+ IR3. The equivalent resistance
of the circuit can be calculated from V = IR. The two equation yield
R = R1 + R2 + R3.
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 R1 = 4 Ω, R2 = 4 Ω, and R3
= 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
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 I1, I2,
and I3 flowing through R1, R2, and R3,
respectively. These three currents recombine when the branches meet again.
I = I1 + I2 + I3.
The voltage across each resistor in the parallel circuit shown is equal to the battery voltage V.
We have I1 = V/R1, I2 = V/R2, I3 = V/R3. We find the equivalent resistance from V = IR,
1/R = I/V = (I1 + I2 + I3)/V = (1/R1) + (1/R2) + (1/R3).
1/R = (1/R1) + (1/R2) + (1/R3).
For a set of parallel resistors the reciprocal of their
equivalent resistance equals the sum of the reciprocals of their individual
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 R1 = 8 Ω, R2
= 8 Ω, and R3 = 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 Ii = V/Ri. The voltage across each resistor is 10 V, therefore
I1 = (10 V)/(8 Ω) = 1.25 A,
I2 = (10 V)/(8 Ω) = 1.25 A ,
I3 = (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.
R1 and R2 are in series and are replaced by R12 = R1 + R2.
Now R12 and R3 are in parallel and can be replaced by R, where 1/R = 1/R12+ 1/R3.
If R1 = R2 = R3 = 1 Ω, then R12 = 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 R3 is I = V/R3, the current flowing through R1 and R2 is V/R12 = 3 A.
What are the largest and smallest resistances you can obtain by connecting a 36 Ω, a 50 Ω, and a 700 Ω resistor together?
To obtain the largest resistance, we have to connect the resistors in series.
Rlargest = (36 + 50 + 700) Ω = 786 Ω.
To obtain the smallest resistance, we have to connect the resistors in parallel.
1/Rsmallest = (1/36 + 1/50 + 1/700) Ω-1 = 0.0492/Ω.
Rsmallest = 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 rules apply to all circuit.
In the circuit to the right, let V1 be a 1 V battery, V2 be a 2 V battery, V3 be a 3 V battery, and let each resistor be a 1 Ω resistor. Pick directions for the currents as shown in the diagram.