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Quick check:

Someone with experience may look at this circuit and assume the power is off at first. (Otherwise the answer is simple, all lamps are off.) Then when the power is applied current will flow through all the bulbs except 5. This is because you place 5 at a balancing point of sorts. In order for current to flow through a load the voltage has to be different at each end. But your design is symmetrical so there is not a voltage difference. Going on, as time passes the capacitor will charge up and the current will stop. All the lights will go out.

Longer way:

You can either anylize this circuit using the Thevenin equivalent or the Norton equivalent. One's just saying the cup is half empty and the other half full. Same thing. Now, the tricky bit is that you have included a capacitor. So (big jump) you need to use differential equations! This MIT (capacitor / resistor) work shop should actually address most if not all your questions.

A word about capacitors:

Let us consider an ideal capacitor. Such a device is made of 2 sheets of metal separated by a thin insulator. If you "push" electrons into one metal sheet (charge it negative) you influence the other metal sheet to push electrons out (charge it positive). It is very easy to charge on an empty capacitor. So the capacitor looks like a short. As time passes, the capacitor gets saturated (gets full). Now it is practically impossible to add more charge. The capacitor now looks like an open. If you remove this ideal capacitor it would retain this charge. You could use it later as a power source if you like. However, in the real world, we can only approach what an ideal capacitor is. So, eventually, the charge will leak away.

Quick check:

Someone with experience may look at this circuit and assume the power is off at first. (Otherwise the answer is simple, all lamps are off.) Then when the power is applied current will flow through all the bulbs except 5. This is because you place 5 at a balancing point of sorts. In order for current to flow through a load the voltage has to be different at each end. But your design is symmetrical so there is not a voltage difference. Going on, as time passes the capacitor will charge up and the current will stop. All the lights will go out.

Longer way:

You can either anylize this circuit using the Thevenin equivalent or the Norton equivalent. One's just saying the cup is half empty and the other half full. Same thing. Now, the tricky bit is that you have included a capacitor. So (big jump) you need to use differential equations! This MIT (capacitor / resistor) work shop should actually address most if not all your questions you have asked above.

A word about capacitors:

Let us consider an ideal capacitor. Such a device is made of 2 sheets of metal separated by a thin insulator. If you "push" electrons into one metal sheet (charge it negative) you influence the other metal sheet to push electrons out (charge it positive). It is very easy to charge on an empty capacitor. So the capacitor looks like a short. As time passes, the capacitor gets saturated (gets full). Now it is practically impossible to add more charge. The capacitor now looks like an open. If you remove this ideal capacitor it would retain this charge. You could use it later as a power source if you like. However, in the real world, we can only approach what an ideal capacitor is. So, eventually, the charge will leak away.

Quick check:

Someone with experience may look at this circuit and assume the power is off at first. Then when the power is applied current will flow through all the bulbs except 5. This is because you place 5 at a balancing point of sorts. In order for current to flow through a load the voltage has to be different at each end. But your design is symmetrical so there is not a voltage difference. Going on, as time passes the capacitor will charge up and the current will stop. All the lights will go out.

Longer way:

You can either anylize this circuit using the Thevenin equivalent or the Norton equivalent. One's just saying the cup is half empty and the other half full. Same thing. Now, the tricky bit is that you have included a capacitor. So (big jump) you need to use differential equations! This MIT (capacitor / resistor) work shop PDF should actually address most if not all your questions.

A word about capacitors:

Let us consider an ideal capacitor, forget about current (the flow of holes (not electrons)) and about voltage. A capacitor is made of 2 sheets of metal separated by a thin insulator. If you "push" electrons into one metal sheet (charge it negative) you influence the other metal sheet to push electrons out (charge it positive). It is very easy to put a charge on a capacitor which it is "empty" to start with. So the capacitor looks like a short. As time goes on, the capacitor gets saturated (gets full). Now it is practically impossible to add more charge. The capacitor now looks like an open. If you remove this ideal capacitor it would retain this charge. You could use it later as a power source if you like. However, in the real world, we can only approach what an ideal capacitor is. So, eventually, the charge will leak away.

Quick check:

Someone with experience may look at this circuit and assume the power is off at first. (Otherwise the answer is simple, all lamps are off.) Then when the power is applied current will flow through all the bulbs except 5. This is because you place 5 at a balancing point of sorts. In order for current to flow through a load the voltage has to be different at each end. But your design is symmetrical so there is not a voltage difference. Going on, as time passes the capacitor will charge up and the current will stop. All the lights will go out.

Longer way:

You can either anylize this circuit using the Thevenin equivalent or the Norton equivalent. One's just saying the cup is half empty and the other half full. Same thing. Now, the tricky bit is that you have included a capacitor. So (big jump) you need to use differential equations! This MIT (capacitor / resistor) work shop should actually address most if not all your questions.

A word about capacitors:

Let us consider an ideal capacitor. Such a device is made of 2 sheets of metal separated by a thin insulator. If you "push" electrons into one metal sheet (charge it negative) you influence the other metal sheet to push electrons out (charge it positive). It is very easy to charge on an empty capacitor. So the capacitor looks like a short. As time passes, the capacitor gets saturated (gets full). Now it is practically impossible to add more charge. The capacitor now looks like an open. If you remove this ideal capacitor it would retain this charge. You could use it later as a power source if you like. However, in the real world, we can only approach what an ideal capacitor is. So, eventually, the charge will leak away.

Quick check: Someone with experience may look at this circuit and assume the power is off at first. Then when the power is applied current will flow through all the bulbs except 5. This is because you place 5 at a balancing point of sorts. In order for current to flow through a load the voltage has to be different at each end. But your design is symmetrical so there is not a voltage difference. Going on, as time passes the capacitor will charge up and the current will stop. All the lights will go out.

Longer way: You can either anylize this circuit using the Thevenin equivalent or the Norton equivalent. One's just saying the cup is half empty and the other half full. Same thing. Now, the tricky bit is that you have an active component. The capacitor. So (big jump) you need to use differential equations! This MIT (capacitor / resistor) work shop PDF should actually address most if not all your questions. But most engineers like to simplify things, so use Phasor Analysis, which generally reduces all the differential equation work to simple algebra.

Quick check:

Someone with experience may look at this circuit and assume the power is off at first. Then when the power is applied current will flow through all the bulbs except 5. This is because you place 5 at a balancing point of sorts. In order for current to flow through a load the voltage has to be different at each end. But your design is symmetrical so there is not a voltage difference. Going on, as time passes the capacitor will charge up and the current will stop. All the lights will go out.

Longer way:

You can either anylize this circuit using the Thevenin equivalent or the Norton equivalent. One's just saying the cup is half empty and the other half full. Same thing. Now, the tricky bit is that you have included a capacitor. So (big jump) you need to use differential equations! This MIT (capacitor / resistor) work shop PDF should actually address most if not all your questions.

A word about capacitors:

Let us consider an ideal capacitor, forget about current (the flow of holes (not electrons)) and about voltage. A capacitor is made of 2 sheets of metal separated by a thin insulator. If you "push" electrons into one metal sheet (charge it negative) you influence the other metal sheet to push electrons out (charge it positive). It is very easy to put a charge on a capacitor which it is "empty" to start with. So the capacitor looks like a short. As time goes on, the capacitor gets saturated (gets full). Now it is practically impossible to add more charge. The capacitor now looks like an open. If you remove this ideal capacitor it would retain this charge. You could use it later as a power source if you like. However, in the real world, we can only approach what an ideal capacitor is. So, eventually, the charge will leak away.