a current of 3 amp is passing through a galvanometer if a high rasistance is connected in series to it the total current that will pass through the resistance will be?

Ans 15 amp

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what is missing bobpursley

To calculate the total current passing through the resistance when a galvanometer of 3 Amps is connected in series with a high resistance, we need to consider the concept of current division.

When multiple resistances are connected in series, the total current passing through the circuit is the same at every point in the circuit.

In this case, the galvanometer and the high resistance are connected in series. Since the total current passing through the circuit is 3 Amps, the same current will flow through the galvanometer.

Now, let's assume the resistance of the high resistance connected in series is R (in ohms).

According to Ohm's Law, the current passing through a resistor (I) is given by the formula: I = V/R, where V is the voltage across the resistor.

Since the total current passing through the circuit is the same, the voltage across the high resistance will be V = I * R.

Substituting the given values, we have V = 3 Amps * R.

If the total current passing through the high resistance is 15 Amps, we can set up the following equation:

15 Amps = 3 Amps * R

Now, we can solve for R:

R = 15 Amps / 3 Amps
R = 5 ohms

Therefore, the resistance of the high resistance connected in series should be 5 ohms to produce a total current of 15 amps passing through it.