
What is Shunt? Obtain formula for the Shunt.
Answer
491.1k+ views
Hint: A galvanometer cannot be used directly to measure big current. To measure large currents, we have to carry out some modifications in the galvanometer. After the modification we get an instrument which is called an ammeter.
Complete step by step solution:
The galvanometer is connected with a small resistance resistor in parallel combination and this resistor is called shunt. The value of shunt is only selected when only the current of the galvanometer passes and the remaining $I-{{I}_{G}}$ through the shunt. Let the resistance of the galvanometer be G. since G and S are in parallel combination, the potential difference across them should be same, i.e.,
${{I}_{G}}G=S(I-{{I}_{G}})$ or $S=\dfrac{G{{I}_{G}}}{I-{{I}_{G}}}$
The equation above gives you the value of shunt to be connected in parallel with the galvanometer to convert into an ammeter of range $0-I$.
An ammeter is used in series to measure the current. The reading of an ammeter is generally lesser than the actual current in the circuit. It is because when we connect an ammeter in the circuit to measure the current, the ammeter introduces its own resistance in the circuit , which results in an increase in the resistance of the circuit and decrease in the current.
Note: The lower the value of shunt resistance, the higher the range of ammeter because current is inversely proportional to resistance the more will be the resistance the less will be the current and the less will be the resistance the more will be the current. As the ammeter is used to measure the current in the circuit so it needs to take a small resistance current in parallel combination.
Complete step by step solution:
The galvanometer is connected with a small resistance resistor in parallel combination and this resistor is called shunt. The value of shunt is only selected when only the current of the galvanometer passes and the remaining $I-{{I}_{G}}$ through the shunt. Let the resistance of the galvanometer be G. since G and S are in parallel combination, the potential difference across them should be same, i.e.,
${{I}_{G}}G=S(I-{{I}_{G}})$ or $S=\dfrac{G{{I}_{G}}}{I-{{I}_{G}}}$
The equation above gives you the value of shunt to be connected in parallel with the galvanometer to convert into an ammeter of range $0-I$.
An ammeter is used in series to measure the current. The reading of an ammeter is generally lesser than the actual current in the circuit. It is because when we connect an ammeter in the circuit to measure the current, the ammeter introduces its own resistance in the circuit , which results in an increase in the resistance of the circuit and decrease in the current.
Note: The lower the value of shunt resistance, the higher the range of ammeter because current is inversely proportional to resistance the more will be the resistance the less will be the current and the less will be the resistance the more will be the current. As the ammeter is used to measure the current in the circuit so it needs to take a small resistance current in parallel combination.
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