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Physics 6B Electric Current And DC Circuit Examples Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

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Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Electric Current is the RATE at which charge flows(usually through a wire). We can define it with a formula: Units are Coulombs/second, or Amperes (A)

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Resistance and Ohm’s Law Under normal circumstances, wires present some resistance to the motion of electrons. Ohm’s law relates the voltage to the current: Be careful – Ohm’s law is not a universal law and is only useful for certain materials (which include most metallic conductors).

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Resistance and Ohm’s Law The units of resistance, volts per ampere, are called ohms: Two wires of the same dimensions will have different resistances if they are made of different materials. This property of a material is called the resistivity.

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Resistance and Ohm’s Law Example: 2 cylindrical resistors are made from the same material. The length of resistor B is twice the length of A, and the radius of B is 3 times the radius of A. The ratio of their resistances, R A /R B, is: a)1/2 b)2/3 c)3/2 d)9/2

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Resistance and Ohm’s Law Example: 2 cylindrical resistors are made from the same material. The length of resistor B is twice the length of A, and the radius of B is 3 times the radius of A. The ratio of their resistances, R A /R B, is: a)1/2 b)2/3 c)3/2 d)9/2

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Energy and Power in Electric Circuits In materials for which Ohm’s law holds, the power can also be written: This power mostly becomes heat inside the resistive material.

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Resistors in Series and Parallel Resistors connected end to end are said to be in series. They can be replaced by a single equivalent resistance without changing the current in the circuit.

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Resistors in Series and Parallel Since the current through the series resistors must be the same in each, and the total potential difference is the sum of the potential differences across each resistor, we find that the equivalent resistance is:

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Resistors in Series and Parallel Resistors are in parallel when they are across the same potential difference; they can again be replaced by a single equivalent resistance:

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Resistors in Series and Parallel Using the fact that the potential difference across each resistor is the same, and the total current is the sum of the currents in each resistor, we find: If you have a pair of resistors in parallel you can use a quick shortcut:

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Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Example: Find the current through, and power used by each resistor in this circuit. Resistances are given as follows: R 1 =6Ω, R 2 =12Ω, R 3 =6Ω 20V R2R2 R1R1 R3R3

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Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 20V 12Ω 6Ω6Ω 6Ω6Ω Example: Find the current through, and power used by each resistor in this circuit. Resistances are given as follows: R 1 =6Ω, R 2 =12Ω, R 3 =6Ω Before we can calculate the individual currents, we need to know how much current is supplied by the battery. So we need to find the equivalent resistance for the circuit.

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Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 20V 12Ω 6Ω6Ω 6Ω6Ω 20V 6Ω6Ω4Ω4Ω Example: Find the current through, and power used by each resistor in this circuit. Resistances are given as follows: R 1 =6Ω, R 2 =12Ω, R 3 =6Ω Parallel – R eq =4Ω Before we can calculate the individual currents, we need to know how much current is supplied by the battery. So we need to find the equivalent resistance for the circuit. Notice that R 1 and R 2 are in parallel. We can combine them into one single resistor: This shortcut formula will work for any pair of parallel resistors.

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Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 20V 12Ω 6Ω6Ω 6Ω6Ω 20V 6Ω6Ω4Ω4Ω 10Ω Example: Find the current through, and power used by each resistor in this circuit. Resistances are given as follows: R 1 =6Ω, R 2 =12Ω, R 3 =6Ω Parallel – R eq =4Ω Series – R eq =10Ω Before we can calculate the individual currents, we need to know how much current is supplied by the battery. So we need to find the equivalent resistance for the circuit. Notice that R 1 and R 2 are in parallel. We can combine them into one single resistor: This shortcut formula will work for any pair of parallel resistors. The next step is to combine the remaining resistors, which are in series. The formula is simple – just add them together.

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Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 20V 12Ω 6Ω6Ω 6Ω6Ω 20V 6Ω6Ω4Ω4Ω 10Ω Example: Find the current through, and power used by each resistor in this circuit. Resistances are given as follows: R 1 =6Ω, R 2 =12Ω, R 3 =6Ω Parallel – R eq =4Ω Series – R eq =10Ω Before we can calculate the individual currents, we need to know how much current is supplied by the battery. So we need to find the equivalent resistance for the circuit. Notice that R 1 and R 2 are in parallel. We can combine them into one single resistor: This shortcut formula will work for any pair of parallel resistors. The next step is to combine the remaining resistors, which are in series. The formula is simple – just add them together. Now that we finally have our circuit simplified down to a single resistor we can use Ohm’s Law to compute the current supplied by the battery: 2 Amps

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Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 20V 12Ω 6Ω6Ω 6Ω6Ω 20V 6Ω6Ω4Ω4Ω 10Ω Example: Find the current through, and power used by each resistor in this circuit. Resistances are given as follows: R 1 =6Ω, R 2 =12Ω, R 3 =6Ω Parallel – R eq =4Ω Series – R eq =10Ω 2 Amps Now that we have the total current, we have to find the individual currents. It might help to make a table like this: Current (I)Voltage (V)Power (P) R 1 =6Ω R 2 =12Ω R 3 =6Ω Battery

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Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 20V 12Ω 6Ω6Ω 6Ω6Ω 20V 6Ω6Ω4Ω4Ω 10Ω Example: Find the current through, and power used by each resistor in this circuit. Resistances are given as follows: R 1 =6Ω, R 2 =12Ω, R 3 =6Ω Parallel – R eq =4Ω Series – R eq =10Ω 2 Amps Now that we have the total current, we have to find the individual currents. It might help to make a table like this: Current (I)Voltage (V)Power (P) R 1 =6Ω R 2 =12Ω R 3 =6Ω2 Amps Battery2 Amps20 volts Take a look at the original circuit. Notice that all the current has to go through R 3. So I 3 = 2 Amps. We can also fill in the information for the battery – we know its voltage and we already found the total current, which has to come from the battery.

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Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 20V 12Ω 6Ω6Ω 6Ω6Ω 20V 6Ω6Ω4Ω4Ω 10Ω Example: Find the current through, and power used by each resistor in this circuit. Resistances are given as follows: R 1 =6Ω, R 2 =12Ω, R 3 =6Ω Parallel – R eq =4Ω Series – R eq =10Ω 2 Amps Now that we have the total current, we have to find the individual currents. It might help to make a table like this: Current (I)Voltage (V)Power (P) R 1 =6Ω R 2 =12Ω R 3 =6Ω2 Amps12 volts Battery2 Amps20 volts Take a look at the original circuit. Notice that all the current has to go through R 3. So I 3 = 2 Amps. We can also fill in the information for the battery – we know its voltage and we already found the total current, which has to come from the battery. Now that we have the Current for resistor #3, we can use Ohm’s Law to find the voltage drop:

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Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 20V 12Ω 6Ω6Ω 6Ω6Ω 20V 6Ω6Ω4Ω4Ω 10Ω Example: Find the current through, and power used by each resistor in this circuit. Resistances are given as follows: R 1 =6Ω, R 2 =12Ω, R 3 =6Ω Parallel – R eq =4Ω Series – R eq =10Ω 2 Amps Now that we have the total current, we have to find the individual currents. It might help to make a table like this: Current (I)Voltage (V)Power (P) R 1 =6Ω R 2 =12Ω R 3 =6Ω2 Amps12 volts Battery2 Amps20 volts The current for R1 and R2 will be easier to find if we calculate the voltage drops first (they have to be the same voltage because they are in parallel – make sure you understand why!).

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Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 20V 12Ω 6Ω6Ω 6Ω6Ω 20V 6Ω6Ω4Ω4Ω 10Ω Example: Find the current through, and power used by each resistor in this circuit. Resistances are given as follows: R 1 =6Ω, R 2 =12Ω, R 3 =6Ω Parallel – R eq =4Ω Series – R eq =10Ω 2 Amps Now that we have the total current, we have to find the individual currents. It might help to make a table like this: Current (I)Voltage (V)Power (P) R 1 =6Ω8 volts R 2 =12Ω8 volts R 3 =6Ω2 Amps12 volts Battery2 Amps20 volts The current for R1 and R2 will be easier to find if we calculate the voltage drops first (they have to be the same voltage because they are in parallel – make sure you understand why!). Either use Ohm’s law (use the 2 nd diagram with the combined resistance of 4Ω), or notice that the total voltage is 20V, and R 3 uses 12V, so there is 8V left over for R 1 or R 2. Now we can use Ohm’s Law again for the individual resistors to find the current through each:

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Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 20V 12Ω 6Ω6Ω 6Ω6Ω 20V 6Ω6Ω4Ω4Ω 10Ω Example: Find the current through, and power used by each resistor in this circuit. Resistances are given as follows: R 1 =6Ω, R 2 =12Ω, R 3 =6Ω Parallel – R eq =4Ω Series – R eq =10Ω 2 Amps Now that we have the total current, we have to find the individual currents. It might help to make a table like this: Current (I)Voltage (V)Power (P) R 1 =6Ω4/3 Amps8 volts R 2 =12Ω2/3 Amps8 volts R 3 =6Ω2 Amps12 volts Battery2 Amps20 volts Total = 2 Amps The current for R1 and R2 will be easier to find if we calculate the voltage drops first (they have to be the same voltage because they are in parallel – make sure you understand why!). Either use Ohm’s law (use the 2 nd diagram with the combined resistance of 4Ω), or notice that the total voltage is 20V, and R 3 uses 12V, so there is 8V left over for R 1 or R 2. Now we can use Ohm’s Law again for the individual resistors to find the current through each:

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Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 20V 12Ω 6Ω6Ω 6Ω6Ω 20V 6Ω6Ω4Ω4Ω 10Ω Example: Find the current through, and power used by each resistor in this circuit. Resistances are given as follows: R 1 =6Ω, R 2 =12Ω, R 3 =6Ω Parallel – R eq =4Ω Series – R eq =10Ω 2 Amps Now that we have the total current, we have to find the individual currents. It might help to make a table like this: Current (I)Voltage (V)Power (P) R 1 =6Ω4/3 Amps8 volts R 2 =12Ω2/3 Amps8 volts R 3 =6Ω2 Amps12 volts Battery2 Amps20 volts Finally, we can calculate the power for each circuit element. You have your choice of formulas:

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Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 20V 12Ω 6Ω6Ω 6Ω6Ω 20V 6Ω6Ω4Ω4Ω 10Ω Example: Find the current through, and power used by each resistor in this circuit. Resistances are given as follows: R 1 =6Ω, R 2 =12Ω, R 3 =6Ω Parallel – R eq =4Ω Series – R eq =10Ω 2 Amps Now that we have the total current, we have to find the individual currents. It might help to make a table like this: Current (I)Voltage (V)Power (P) R 1 =6Ω4/3 Amps8 volts32/3 Watts R 2 =12Ω2/3 Amps8 volts16/3 Watts R 3 =6Ω2 Amps12 volts24 Watts Battery2 Amps20 volts40 Watts Finally, we can calculate the power for each circuit element. You have your choice of formulas: I suggest using the simplest one. Plus it’s easy to remember because you probably live there… As a final check you can add the powers to make sure they come out to the total power supplied by the battery.

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Kirchhoff’s Rules More complex circuits cannot be broken down into series and parallel pieces. For these circuits, Kirchhoff’s rules are useful. The junction rule is a consequence of charge conservation; the loop rule is a consequence of energy conservation.

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Kirchhoff’s Rules The junction rule: At any junction, the current entering the junction must equal the current leaving it.

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Kirchhoff’s Rules The loop rule: The algebraic sum of the potential differences around a closed loop must be zero (it must return to its original value at the original point).

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Kirchhoff’s Rules Using Kirchhoff’s rules: The variables for which you are solving are the currents through the resistors. You need as many independent equations as you have variables to solve for. You will need both loop and junction rules.

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Example – Find the current through each resistor, and find the power used by each.

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Example – Find the current through each resistor. We can start by writing a formula for each loop shown. The direction of current has already been chosen for us, but that is an important first step.

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Example – Find the current through each resistor. Loop 1: +15V -100I 3 -100I 1 = 0 Loop 2: -9V -100I 2 +100I 3 = 0

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We can start by writing a formula for each loop shown. The direction of current has already been chosen for us, but that is an important first step. Example – Find the current through each resistor. Loop 1: +15V -100I 3 -100I 1 = 0 Loop 2: -9V -100I 2 +100I 3 = 0 We also need to write down a junction formula: At point A we have +I 1 -I 2 -I 3 = 0

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Example – Find the current through each resistor. We can start by writing a formula for each loop shown. The direction of current has already been chosen for us, but that is an important first step. The rest is just algebra to solve for the 3 unknown currents. Loop 1: +15V -100I 3 -100I 1 = 0 Loop 2: -9V -100I 2 +100I 3 = 0 We also need to write down a junction formula: At point A we have +I 1 -I 2 -I 3 = 0

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Example – Find the current through each resistor. We can start by writing a formula for each loop shown. The direction of current has already been chosen for us, but that is an important first step. Loop 1: +15V -100I 3 -100I 1 = 0 Loop 2: -9V -100I 2 +100I 3 = 0 We also need to write down a junction formula: At point A we have +I 1 -I 2 -I 3 = 0 The rest is just algebra to solve for the 3 unknown currents. Answers: I 1 =0.07A, I 2 =-0.01A, I 3 =0.08A Notice that I 2 came out negative. This is no problem – it just means that the current actually flows the opposite way from what was chosen in the picture.

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Circuits Containing Capacitors Capacitors can also be connected in series or in parallel. When capacitors are connected in parallel, the potential difference across each one is the same.

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Circuits Containing Capacitors Therefore, the equivalent capacitance is the sum of the individual capacitances:

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Circuits Containing Capacitors Capacitors connected in series do not have the same potential difference across them, but they do all carry the same charge. The total potential difference is the sum of the potential differences across each one.

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Circuits Containing Capacitors Therefore, the equivalent capacitance is Capacitors in series combine like resistors in parallel, and vice versa. Note that this equation gives you the inverse of the capacitance, not the capacitance itself!

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RC Circuits In a circuit containing only batteries and capacitors, charge appears almost instantaneously on the capacitors when the circuit is connected. However, if the circuit contains resistors as well, this is not the case.

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RC Circuits Using calculus, it can be shown that the charge on the capacitor increases as: Here, τ is the time constant of the circuit: And is the final charge on the capacitor, Q.

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RC Circuits Here is the charge vs. time for an RC circuit:

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RC Circuits It can be shown that the current in the circuit has a related behavior:

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RC Circuits Interactive example: Charging a capacitor

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