Circuit Exercises: | 1A | 1B | 1C | 2A | 2B | 2CGo up
 - by Kelly Intile, 2003

1A.

First, find the total resistance by adding up all the resistors.  8 + 4 = 12 Ohms.

Remember that series circuits have only one current, but many voltage drops, so find the one current I = V/R = 24/12 = 2 Amps.  This is the voltage in A1.  

Now, use this current and each of the resistors to find the voltage in V1 and V2 with the formula V = IR:

For V1:   Resistance = 8 Ohms, Current = 2 Amps.    V = IR = 2*8 = 16 V.

For V2:   Resistance = 4 Ohms, Current = 2 Amps.    V = IR = 2*4 = 8 V.

To find the power at the 4 Ohm resistor, use the formula P = I*R  (2) * 4 = 16 W.
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1B. 

This is also a series circuit, so find the total resistance.  3.2+6.5+5.2+7.5 = 22.4 Ohms.  Now that you know the resistance, you can use that and the voltage to find the one current.  I = V/R = 12/22.4 = .536 A.  This is the current on both A1 and A2.

Now we have to find the voltage at V1, V2, and V3.  Use the one current (.536 A) and each respective resistor to find voltage.  The only difference between this circuit and the previous one is that some of these voltage drops cover more than one resistor.  Just add the resistors up and multiply them by the current to find the voltage.

At V1:  Resistance = 6.5 + 5.2 = 11.7 Ohms.  Current = .536 A.  V = IR = 11.7*.536 = 6.268 V

At V2:  Resistance = 3.2 Ohms.  Current = .536 A.  V = IR = 3.2*.536 = 1.714 V

At V3:  Resistance = 7.5 + 5.2 = 12.7 Ohms.  Current = .536 A.  V = IR = 12.7*.536 = 6.804 V

To find the power at the 6.5 Ohm resistor, use the formula P = I*R  (.536) * 6.5 = 1.865 W
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1C.

 

Find the total resistance first by adding up all the resistors.  13.5+16.8+7.27+18+6.5+8.5 = 70.57 Ohms.  Now, find the one current (since this is a series circuit, there is only one current).  I = V/R = 47/70.57 = .666 A.  This current is the reading on both A1 and A2.

Find the voltages by adding up whatever resistors they cover and multiplying that by the current (V = IR).

For V1:  Resistance = 7.27+18 = 25.27 Ohms, Current = .666 A.  V = IR = .666*25.27 = 16.830 V.

For V2:  Resistance = 8.5 Ohms, Current = .666 A.  V = IR = .666*8.5 = 5.661 V.

For V3:  Resistance = 7.27+18+6.5+8.5= 40.27 Ohms, Current = .666 A.  V = IR = .666*40.27 = 26.820 V.

The power lost at the 13.5 Ohm resistor can be found using the formula P = I*R  P = (.666) * 13.5 = 6.0 W. 

 

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2A.

Parallel circuits have many different currents, but only one voltage drop.  First, find the current that goes through A2, A3, and then A1 (we will call this I1).  By adding up all the resistance on that current (R = 4 Ohms) and using the given voltage (V = 12 V), find the current using I = V/R.  I = 12/4 = 3 Amps.  Don't assume that this current is the reading on A1, A2, and A3, because some of those ammeters have more than one current going through them.

Now find the second current using the same method.  This time, the resistance is 3 Ohms, and the voltage is still 12 Volts.  I = V/R = 12/3 = 4 Amps.  The third current (I3) is the same, and since its resistance is 2 Ohms, its current is 12/2 Amps (6 Amps).

Now that we have all three currents, we can find all the ammeters.  As you can see from the diagram, all three currents go through A1, so the current at A1 is found by adding up all three currents.  I1 + I2  + I3 = 3 + 4 + 6 = 13 Amps, which is the same as A2 because A2 also has all three currents running through it.

A3 only has the first current running through it, so its reading is 3 Amps. A4 has both I2 and I3, so its reading is I2 + I3 = 4 + 6 = 10 Amps.

A5 and A6 both only read the third current, so their reading the same as I3, 6 Amps.

Now we need to find the voltage drops.  The resistance on V1 is 0 Ohms, so the voltage drop is 0 V.  The resistance at V2 is 2 Ohms, and the current you use is the current running through it (I3 = 6 A).  So V = IR = 2*6 = 12 V.  V3 is the same, only this time you use current I2 (4 A) and the resistance is 3 Ohms. V = IR = 4*3 = 12 V.

You also need to find the power at the 3 Ohm resistor, so use the formula P = I*R.  (Use I2, which is 4 A).  P = 16*3 = 48 W.
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2B.

Find the three currents on this circuit just like you did before.  Don't let the triangles mess you up because it's exactly the same.  I1 has the given voltage of 15 V and a resistance of 7 Ohms, so I1 = V/R = 15/7 = 2.14 Amps.  I2 has the same voltage and an 8 Ohm resistance, so I2 = V/R = 15/8 = 1.88 Amps.  I3 is the same except its resistance is 9 Ohms, so I3 = V/R = 15/9 = 1.67 Amps.

To find the readings on the ammeters, decide which of the three currents are flowing through them.  A1 has all three currents, so its reading is I1 + I2 + I3 = 2.14+1.875+1.67 = 5.68 Amps.  I2 and I3 both flow through A2, so A2 = I2 + I3 = 1.875 + 1.67 = 3.54 Amps.  Current I3 is the only one running through A3, so A3 reads 1.67 Amps.

The voltage drops are found using the same process as the last circuit (again, try to ignore the triangular shape).  V1 covers no resistance, so its R = 0.  Therefore, V1 = 0 V.  V2 has a 9 Ohm resistor, and, using current I3 (1.67 A), V = IR = 9*1.67 = 15 V.
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2C.

First find the three currents in this circuit.  I1 is found using I = V/R with V = 48 Volts and R = 13 Ohms.  I = 48/13 = 3.692 Amps. I2 has two resistors, so you need to add them up to get the total resistance.  17 + 19 = 36.  I2 = 48/36 = 1.333 Amps.  The third current (I3) is found the same way as I1, except this time with a resistance of 37 Ohms.  I3 = 48/37 = 1.297 Amps.

Now that we know the currents, we can start finding the readings on the ammeters.  A1 has all three currents flowing through it, so add up all the currents to find its reading.  A1 = 3.692+1.333+1.297 = 6.323 Amps.  A2 has the second and third currents running through it, so its reading is I2 + I3 = 1.333+1.297 = 2.630 Amps.  Current I2 is the only one read by A3, so A3 = I2 = 1.333 Amps.  A4 is the same, except it only reads current I3, so A4 = I3 = 1.297 Amps.

Now we find the voltage drop.  V1 covers a 17 Ohm resistor and its voltage is found by V = IR.  The current flowing through is I2 (1.333 A), so V = IR = 17*1.333 = 22.66 V.
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