The Magic of the Galvanometer
Imagine you have a highly sensitive instrument—a microammeter—that twitches its needle at the slightest whisper of electrical current. In our case, this galvanometer has a resistance of 100 Ω and reaches its maximum limit (full-scale deflection) at a mere 50 μA.
But what if you want to measure a roaring river of current, like 5 mA, or a substantial potential difference, like 10 V? You don't need to buy a new instrument; you just need to cleverly redirect the electricity. This is the beautiful art of shunts and multipliers.
The Ammeter Transformation
Bypassing the Flood
To convert our delicate galvanometer into a robust ammeter, we must protect it from the excess current. We do this by providing an alternate, low-resistance path—a shunt (S)—connected in parallel.
When a large current
i enters the setup, only the safe amount
ig goes through the galvanometer, while the rest
(i−ig) rushes through the shunt. Because they are in parallel, the voltage drop across both paths is identical:
Vshunt=Vgalvanometer
S(i−ig)=igG
Rearranging this gives us our master equation for the shunt resistance:
S=i−igigG
Let's test
Option (c), which proposes a
5 mA range. Substituting our values:
S=5×10−3−50×10−650×10−6×100
Here is a pro-tip:
50 μA is incredibly small compared to
5 mA (
5000 μA). We can safely approximate the denominator to just
5000 μA.
S≈5000×10−65000×10−6=1 Ω
A 1 Ω parallel resistance perfectly converts our instrument into a 5 mA ammeter. Option (c) is correct!
The Voltmeter Transformation
Dropping the Pressure
Now, let's switch gears. What if we want to measure voltage? A voltmeter must be connected in parallel across a component, which means it will experience the full voltage. To prevent our sensitive galvanometer from frying, we must connect a massive resistance—a multiplier (R)—in series.
This high resistance ensures that even at the maximum voltage
V, only the tiny full-scale current
ig flows through the instrument. The total voltage is the sum of the voltage drops:
V=ig(G+R)
Rearranging for
R, we get:
R=igV−G
Let's evaluate
Option (b), which suggests a
10 V range. Plugging in the numbers:
R=50×10−610−100
R=200,000−100=199,900 Ω
In the practical world of electronics, 100 Ω is completely negligible when standing next to 200,000 Ω. We can confidently round this to 200 kΩ. Option (b) is also correct!
Final Conclusion
By mastering the principles of parallel and series circuits, we've unlocked the true versatility of the galvanometer. A tiny 1 Ω shunt turns it into a 5 mA ammeter, while a massive 200 kΩ series resistor transforms it into a 10 V voltmeter.
Correct Options: (b) and (c)