Understanding DC Circuit Diagrams - AP Physics 2

Our first step in solving this circuit is to find the equivalent resistance of and . If you have two elements in parallel (whether they be resistors, capacitors, or inductors), an easy way to do this is

This is the same as the following equation, which may look more familiar; however, the first equation only works for two elements.

Now that we have the equivalent of the two resistances, we can combine it in series with to get the total .

Now that we have the total resistance, we can combine it with the total voltage given in the problem and use Ohm's law to calculate the current.

The key to solving this problem is to recognize the structure of the circuit. The voltage source, the capacitor, and the two resistors are in parallel to each other, meaning that they have the same voltage! This is a very important concept and will make many problems easier to solve.

Since the branch containing the two resistors has 5V, we can easily solve for the voltage of .

One easy way to solve for the voltage is to calculate the current going through the entire branch. Keep in mind here that two resistors in series have the same current.

Now that we have the current that flows through each of the two resistors, we can calculate the voltage drop across only the first resistor.

The first step to solving this problem is to try to simplify it. We can combine resistors and into one equivalent resistance:

Now, we have a single loop with multiple circuit components all connected in series. To calculate the current going through this loop, we need to use Kirchoff's loop law, which says that the sum of all the voltages in a single loop is equal to 0.

This may look confusing at first. Starting from the bottom left point in the circuit, let's begin to go around the loop. The voltage on the battery is +12. The voltage of the resistor is from Ohm's law and since voltage is lost, it's negative.

Continuing on to the capacitor, the voltage across it is , and is negative because voltage is lost across it.

Finally, on to the equivalent resistor, the voltage across it is .

Plugging in all the numbers from the problem and solving the equation above for gives us

Our first step in solving this circuit is to find the equivalent resistance of and . If you have two elements in parallel (whether they be resistors, capacitors, or inductors), an easy way to do this is

This is the same as the following equation, which may look more familiar; however, the first equation only works for two elements.

Now that we have the equivalent of the two resistances, we can combine it in series with to get the total .

Now that we have the total resistance, we can combine it with the total voltage given in the problem and use Ohm's law to calculate the current.

The key to solving this problem is to recognize the structure of the circuit. The voltage source, the capacitor, and the two resistors are in parallel to each other, meaning that they have the same voltage! This is a very important concept and will make many problems easier to solve.

Since the branch containing the two resistors has 5V, we can easily solve for the voltage of .

One easy way to solve for the voltage is to calculate the current going through the entire branch. Keep in mind here that two resistors in series have the same current.

Now that we have the current that flows through each of the two resistors, we can calculate the voltage drop across only the first resistor.

The first step to solving this problem is to try to simplify it. We can combine resistors and into one equivalent resistance:

Now, we have a single loop with multiple circuit components all connected in series. To calculate the current going through this loop, we need to use Kirchoff's loop law, which says that the sum of all the voltages in a single loop is equal to 0.

This may look confusing at first. Starting from the bottom left point in the circuit, let's begin to go around the loop. The voltage on the battery is +12. The voltage of the resistor is from Ohm's law and since voltage is lost, it's negative.

Continuing on to the capacitor, the voltage across it is , and is negative because voltage is lost across it.

Finally, on to the equivalent resistor, the voltage across it is .

Plugging in all the numbers from the problem and solving the equation above for gives us

A represents a battery. Convention dictates that the direction of charge flow is from the small side to the large side. B is a resistor. Its symbol makes sense if you view the flow of electrons similarly to water flowing; water's flow is resisted by turns, so the jagged lines would evoke that thought for electrons flowing. C is a capacitor. Common capacitors are parallel plates of equal surface area separated by a vacuum or dielectric. This is why the lines are equal in length, unlike the battery. D is a solenoid. A solenoid is a coil of wire that induces a magnetic inside of it.

Our first step in solving this circuit is to find the equivalent resistance of and . If you have two elements in parallel (whether they be resistors, capacitors, or inductors), an easy way to do this is

This is the same as the following equation, which may look more familiar; however, the first equation only works for two elements.

Now that we have the equivalent of the two resistances, we can combine it in series with to get the total .

Now that we have the total resistance, we can combine it with the total voltage given in the problem and use Ohm's law to calculate the current.

The key to solving this problem is to recognize the structure of the circuit. The voltage source, the capacitor, and the two resistors are in parallel to each other, meaning that they have the same voltage! This is a very important concept and will make many problems easier to solve.

Since the branch containing the two resistors has 5V, we can easily solve for the voltage of .

One easy way to solve for the voltage is to calculate the current going through the entire branch. Keep in mind here that two resistors in series have the same current.

Now that we have the current that flows through each of the two resistors, we can calculate the voltage drop across only the first resistor.

The first step to solving this problem is to try to simplify it. We can combine resistors and into one equivalent resistance:

Now, we have a single loop with multiple circuit components all connected in series. To calculate the current going through this loop, we need to use Kirchoff's loop law, which says that the sum of all the voltages in a single loop is equal to 0.

This may look confusing at first. Starting from the bottom left point in the circuit, let's begin to go around the loop. The voltage on the battery is +12. The voltage of the resistor is from Ohm's law and since voltage is lost, it's negative.

Continuing on to the capacitor, the voltage across it is , and is negative because voltage is lost across it.

Finally, on to the equivalent resistor, the voltage across it is .

Plugging in all the numbers from the problem and solving the equation above for gives us

A represents a battery. Convention dictates that the direction of charge flow is from the small side to the large side. B is a resistor. Its symbol makes sense if you view the flow of electrons similarly to water flowing; water's flow is resisted by turns, so the jagged lines would evoke that thought for electrons flowing. C is a capacitor. Common capacitors are parallel plates of equal surface area separated by a vacuum or dielectric. This is why the lines are equal in length, unlike the battery. D is a solenoid. A solenoid is a coil of wire that induces a magnetic inside of it.

A represents a battery. Convention dictates that the direction of charge flow is from the small side to the large side. B is a resistor. Its symbol makes sense if you view the flow of electrons similarly to water flowing; water's flow is resisted by turns, so the jagged lines would evoke that thought for electrons flowing. C is a capacitor. Common capacitors are parallel plates of equal surface area separated by a vacuum or dielectric. This is why the lines are equal in length, unlike the battery. D is a solenoid. A solenoid is a coil of wire that induces a magnetic inside of it.

Our first step in solving this circuit is to find the equivalent resistance of and . If you have two elements in parallel (whether they be resistors, capacitors, or inductors), an easy way to do this is

This is the same as the following equation, which may look more familiar; however, the first equation only works for two elements.

Now that we have the equivalent of the two resistances, we can combine it in series with to get the total .

Now that we have the total resistance, we can combine it with the total voltage given in the problem and use Ohm's law to calculate the current.

The key to solving this problem is to recognize the structure of the circuit. The voltage source, the capacitor, and the two resistors are in parallel to each other, meaning that they have the same voltage! This is a very important concept and will make many problems easier to solve.

Since the branch containing the two resistors has 5V, we can easily solve for the voltage of .

One easy way to solve for the voltage is to calculate the current going through the entire branch. Keep in mind here that two resistors in series have the same current.

Now that we have the current that flows through each of the two resistors, we can calculate the voltage drop across only the first resistor.

The first step to solving this problem is to try to simplify it. We can combine resistors and into one equivalent resistance:

Now, we have a single loop with multiple circuit components all connected in series. To calculate the current going through this loop, we need to use Kirchoff's loop law, which says that the sum of all the voltages in a single loop is equal to 0.

This may look confusing at first. Starting from the bottom left point in the circuit, let's begin to go around the loop. The voltage on the battery is +12. The voltage of the resistor is from Ohm's law and since voltage is lost, it's negative.

Continuing on to the capacitor, the voltage across it is , and is negative because voltage is lost across it.

Finally, on to the equivalent resistor, the voltage across it is .

Plugging in all the numbers from the problem and solving the equation above for gives us