Final answer:
When a high resistance is added in series with a galvanometer through which 3 amperes are passing, the total current through the resistance would be less than 3 amperes, as the series connection does not increase the current.
Step-by-step explanation:
You've asked, "If a high resistance is connected in series with a galvanometer already carrying a current of 3 amperes, what would be the total current that will pass through the resistance?" The answer is c. Less than 3 A.
When discussing electrical circuits and their components, the concept of series and parallel connections must be clearly understood. In a series circuit, current flows through each component one after the other. This means that the same current flows through all components because there is only one path for the current to flow. Therefore, when a high resistance is added in series to an existing circuit that has a galvanometer, the total current through the circuit remains the same.
However, because you are adding more resistance, Ohm's law (V=IR) tells us that for a given voltage, increasing the resistance results in a lower current. As such, if you begin with 3 amperes passing through the galvanometer without the added high resistance, this value represents the maximum current that the circuit will carry. Adding more resistance cannot increase the current; it will either stay the same or decrease. Since we know the resistance is increased, the current through the high resistance hence the entire series circuit will be less than 3 amperes.