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7

There IS a potential, and all three bulbs will be on. First, you need a reference point to measure voltages. Let's take the wire right of the right battery, and define its potential as 0V. If both batteries generate a voltage of 5V, the wire between the batteries will have a potential of 5V, and the wire left of the batteries will have 10V: So, bulb 1 ...

3

The ammeter measure the current flowing through itself. If you want to measure the current flowing through another component, then you must make the current through the ammeter equal to the current through the component. If you wire it in series, that's true. If you wired it in parallel, the current would be unevenly divided between the component and the ...

1

C2 and C6 are AC coupling capacitors. An AC signal on their input appears on the output. If you just connected these two capacitors together, they would act as a voltage divider - the voltage at the connecting node is the mean of the voltages at the other side of the capacitors. The "weak coupling" comes about from the fact that since the point connecting ...

1

I'm assuming $V_R$ is taken to be the voltage across the resistor in a series RC circuit. The transfer function comes directly for the voltage division rule: $$\frac{V_R}{V_\text{in}} = \frac{Z_{R}}{Z_\text{series}} = \frac{R}{R+\frac{1}{i\omega C}} \, .$$ In this equation $V_R$ and $V_\text{in}$ are phasors, meaning that the actual time dependent ...

1

Bulb H is connected to second battery directly. As long as this bulb is not short circuited, there will be some applied potential difference across its terminals and hence it will be ON

1

Point particles as the electrons (which are the charge carriers) move according to Newton's law $\textbf{F}=q\textbf{E}=m\textbf{a}$. Whenever an electric field is present it generates a difference of potential between two points $A$ and $B$ given by its differential form calculated between the two points  V_A - V_B = \int_A^B \textbf{E}\cdot d\textbf{s}. ...

1

Since the circuit is fully symmetrical (left/right symmetry) the potential at C is exactly half the potential between A and B. This means that there is no current flowing across the point (from the "tip of the V" to the middle of the two resistors at point C), and you can break it without changing the underlying equations describing the current flow.

1

There is too much confusion here to answer meaningfully. Current always flows through the battery. What goes in one end comes out the other. Some types of batteries can be charged by forcing current thru them backwards. Those are called rechargable batteries. Others don't work that way and can be damaged by reverse current. Those are often called ...

1

How does an inductor store [electro]magnetic energy? Rather surprisingly, it's something like a flywheel. You can see a mention of that here in Daniel Reynolds' electronics course: . It really is like this, check out the pictures of inductors on Wikipedia, and you'll notice they're rather like a solenoid. And there's the flywheel again: "As a result, ...

1

The energy is stored in the magnetic field. I usually think of it as "magnetic field lines repel" but that is not very precise (useful for intuition though). But along the same lines as your capacitor example (moving the plates to infinity takes work), if you look at a simple current loop there is a force on the wires from the magnetic field generated. This ...

1

There are broadly three classes of materials: conductor, semiconductor, insulator. The conductor contains a LOT of electrons per unit volume. If you were to charge it, you would add a few more electrons. How many? Let's take copper. It has roughly $8.5\cdot 10^{28}$ electrons per $m^3$. If you have a wire of radius $r$ the number of electrons scales with ...

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