Question

The deflection in the galvanometer falls from 50 divisions to 20 when a 12 ohm shunt is applied. The galvanometer resistance is given by

Answer 1

The 12 ohm shunt would now be carrying 30 divisions worth of current (50 - 20). You did not give the value of the full scale reading or if it was a 50 ma or a 100 ma meter. Lets make an assumption and assume it is a 50 ma galvanometer. We now would employ Ohms Law.

The voltage across the meter and shunt would be 30ma times 12 Ohm or .03 times 12 or .36 volts (E=IR). The resistance of the meter would be (using Ohms Law) E=IR or .36 volts = 20 ma R
R = .36/.02 = 18 Ohms.

There are other ways to solve, but this demonstrates how important Ohms Law is in electronics. The answer would of still been 18 Ohms regardless if it was a microamp galvanometer or a milliamp galvanometer.

Answer 2

i can not get your point plzz help me with wirtten eqvation plzz

Answer 3

An understanding of the concepts would help.
A meter shunt allows a meter to measure current that exceeds the meter's is rated or full scale reading. The shunt is placed in parallel with the meter allowing the additional current to be "shunted" around the meter rather than through the meter. The meter also has an internal resistance. This is what the problem wants you to determine.
Now think about this..........if the shunt resistor had equaled the meter's internal resistance, half of the current would have traveled thru the meter and the other half thru the shunt resistor, and the meter would have gone from 50 divisions to 25 divisions.

But in the problem the meter went from 50 divisions to 20 (a decrease of 30......more than half). What does this tell you? It should indicate to you that the shunt resistor is less than 1/2 the resistance of the meter as more than 1/2 of the current was deflected thru the shunt.....think about it.
Since you now should know that the meter is indicating 20 when prior to applying the shunt it was indicating 50 divisions that there is now 30 divisions of current flowing through the 12 Ohm shunt and 20 divisions of current thru the meter (given by the problem) You now can calculate the voltage across the meter (Ohms Law noting that the voltage across the shunt is also the voltage across the meter). Do the math using Ohms Law.