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A  galvanometer of resistance 40Ω gives a deflection of 5 divisions per mA. There are 50 divisions on the scale. The maximum current that can pass through it when a shunt resistance of 2Ω is connected is
  • a)
    210 mA
  • b)
    155 mA
  • c)
    420 mA
  • d)
    75 mA
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
A galvanometer of resistance 40Ω gives a deflection of 5 divisio...
IG = 50/5 = 10mA; R= 40Ω, Rs = 2Ω
Maximum current,
I = 

= 210 mA
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Most Upvoted Answer
A galvanometer of resistance 40Ω gives a deflection of 5 divisio...
Ohm is connected in series with a resistance of 60 Ohm. The combination is connected across a potential difference of 12 V. Calculate the value of the current passing through the galvanometer.

To find the current passing through the galvanometer, we can use the concept of current division in a series circuit.

First, let's calculate the total resistance in the circuit:
Total resistance = Resistance of galvanometer + Resistance in series
Total resistance = 40 Ohm + 60 Ohm
Total resistance = 100 Ohm

Next, we can use Ohm's Law to find the current passing through the circuit:
Current (I) = Voltage (V) / Total resistance (R)
Current (I) = 12 V / 100 Ohm
Current (I) = 0.12 A

Therefore, the value of the current passing through the galvanometer is 0.12 A.
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