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The simplest electric circuit contains a source of electric energy such as a dry cell or wall outlet, a load such as a light bulb or fan, and conductors such as wires to provide a path for electric current to flow. The resistor, a load that converts electric energy to heat, will be used in all diagrams to represent the load. A dry cell will be used to represent the energy source.
When water flows through a hose friction between the water and the hose impedes the flow. Pinching the hose further adds more friction and reduces the flow more. Water pressure is what is needed to keep the water moving. The more pressure, the more water flows; the more friction, the less water flows.
As current flows through the conductor, the atomic structure of the conductor provides resistance to current flow in much the same way a hose impedes water flow. The more energy the source provides, the more current flows. With more resistance, less current flows. Unlike a water hose, an electric circuit has to make a complete circle from the source through the load(s) and back to the source again or no current will flow.
Resistance, R, is measured in ohms; current, I, (the amount of charge that flows with time) is measured in amperes (A), and the energy provided per coulomb of charge provided by the source is measured in volts (V). All resistance will be attributed to the load(s); the resistance of the conductors and the internal resistance of the power source will be considered negligible.
The relationship between these quantities is given by Ohm's Law:
V = IR
In series circuits, all current goes through every load; there are no forks or branches in the circuit. So, the current, I1, through resistor, R1, and the current, I2, through resistor R2 are the same and equal to the current, I, of the circuit. Just like pinching two places in the same hose results in less flow then pinching in only one place, adding resistors in series increases the total resistance of the circuit, and reduces current.
In accordance with the Law of Conservation of energy, the voltage drops across each resistor (the energy used by each resistor) add up to the voltage (energy) supplied by the source.
In summary, for series circuits,
ITOT = I1 = I2 = . . . .
RTOT = R1 + R2 + . . . .
VTOT = V1 + V2 + . . . .
2. In the diagram of a series circuit above, suppose R1 and R2 were light bulbs. If the light bulb at R2 burned out, what would happen to the brightness of the light bulb at R1?
3. Two resistors, 2.0 ohm and 3.0 ohm, are connected in series with a 12 V battery. Calculate:
(a) the total resistance of the circuit;
(b) the current leaving the battery;
(c) the current through each resistor; (d) the voltage drop across each resistor.
In parallel circuits not all of the current goes through every load. There is at least one fork or branch in the circuit. In the diagram below, the current, ITOT, that leaves the source is split. Some current, I1, goes through R1 and the rest, I2 goes through R2 . The sum of the currents I1 and I2 equals the total current ITOT.
Unlike the series circuit, adding more branches reduces resistance and increases current. This is easy to understand: Suppose a large city developed on both banks of a river and there was only one bridge across the river. If the bridge was narrow, a traffic jam would probably develop at least during rush hour. One way to reduce the resistance to traffic flow would be to build another bridge. With more bridges across the river, more traffic flows. With more branches in the circuit more current can flow, and the less resistance there is to current flow.
The voltage supplied to each branch is the same as the voltage supplied to the source. A branch with low resistance carries more current than a branch with high resistance.
In summary, for parallel circuits,
ITOT = I1 + I2 + . . . .
VTOT = V1 = V2 = . . . .
1/(REFF) = 1/(R1) + 1/(R2) + . . .
4. In the diagram of a parallel circuit above, suppose R1 and R2 were light bulbs. If the light bulb at R2 burned out, what would happen to the brightness of the light bulb at R1?
5. Two resistors, 2.0 ohm and 3.0 ohm, are connected in parallel with a 12 V battery. Calculate: (a) the effective resistance of the circuit; (b) the current leaving the battery; (c) the current through each resistor; (d) the voltage drop across each resistor.
6. In this diagram the three resistances each represent identical light bulbs. Describe what happens to the brightness of the other light bulbs when one of them goes out.
7. Sketch a diagram showing a complete circuit involving only a dry cell, one wire, and a flashlight bulb.TOP OF PAGE
In AC circuits containing a coil, self inductance in the coil impedes current in addition to resistance. The relationship between voltage, current and total impedance in an AC circuit resembles Ohm's law:
V = IZ, where Z is the total impedance.
8. If you wished to make a lamp dimmer with a coil of a variable inductance, how might you accomplish this? If you switched with DC would it work?
9. A high-resistance inductor connected to a 12V battery carries 3 A of current. When connected to an ac source with an rms output of 12v and a frequency of 60Hz, the current drops to 2 A. What is the impedance at 60Hz?
10. A 2-V battery has 0.02 ohms internal resistance. Draw a schematic diagram showing a voltmeter and an ammeter being used to measure the battery’s current and voltage. Calculate the values these devices would show.
11. Two parallel resistors (15 and 30 Ohms) are connected in series with a 9 V battery and a 20 Ohm resistor. What is the current through the 15 Ohm resistor?
12. Kirchoff's Law problem (related search terms: kirchoff kirchoffs law) What is the current through the 12-ohm resistor in the circuit to the left?
13. A material of resistivity 370000 Ohms*m is shaped into a hollow cylindrical shell of length 3 cm, inner radius 0.51 cm and outer radius 1.02 cm. A potential difference is applied across the ends of the cylinder. Find the resistance of the cylinder.
14. Two 120 V light bulbs are rated at 75 W and 150 W. Which bulb has a filament with greater resistance?
15. A combination of two parallel resistors of 12.0 ohms and 6.00 ohms is connected in series with a 6.25 ohm resistor and a battery. The battery has an internal resistance of .250 ohms. If the current in the 6.00 resistor is 8.800 A, what is the EMF for the circuit?
16. Power is provided to a lamp 10 hours every day. At 15 cents per kilowatt-hour, how much money would be saved in thirty days by using a 35-watt bulb instead of 150-watt bulb?
17. What is the current in an AC circuit with 120-volts and a 60-watt light bulb?TOP OF PAGE
Selected solutions are printed below.
For solutions to all the problems on this page click here.
1. V = IR
12 = (2.0)R
R = 6 ohms
2. The bulb at R1 would go out because the circuit would be incomplete.