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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 ...


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

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 ...



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