The range of the ammeter is 5 amperes, and the full-scale deflection current is 0.5 micro-amperes. If the resistance of the galvanometer is 50 ohms, then find the shunt resistance.
Answer
Verified
470.1k+ views
Hint: In this question, we need to determine the value of the shunt resistance that should be attached in parallel with the 50 ohms resistance such that the full-scale deflection of the ammeter is 0.5 micro-amperes whereas the total range of the ammeter is 5 Amperes. For this, we will use the relations for Kirchhoff’s law.
Complete step by step answer:The internal resistance of the ammeter is connected in series, and to extend the range of the ammeter, an additional resistance should be connected in parallel with the internal resistance. The resistances connected in parallel are known as Shunt resistances.
Here, the full-scale deflection is $0.5\mu A$ , and the range is 5 Amperes so, the current passing through the shunt resistance is $\left( {5 - 0.5 \times {{10}^{ - 6}}} \right) = 4.9999995A$
As we know, the potential drop across the resistances connected in parallel is the same. SO,
$
{V_g} = {V_{sh}} \\
\left( {0.5\mu A} \right)\left( {50} \right) = \left( {4.9999995} \right)\left( {{R_{sh}}} \right) \\
{R_{sh}} = \dfrac{{50 \times 0.5 \times {{10}^{ - 6}}}}{{4.9999995}} \\
= 5\mu \Omega \\
$
Hence, the value of the shunt resistance that should be connected in parallel with the ammeter is 5 micro-ohms.
Note:It is interesting to note here that as we need to increase the range of the ammeter so we have connected a very low value of the resistance so that a major part of the current inline will flow through it without harming the meter.
Complete step by step answer:The internal resistance of the ammeter is connected in series, and to extend the range of the ammeter, an additional resistance should be connected in parallel with the internal resistance. The resistances connected in parallel are known as Shunt resistances.
Here, the full-scale deflection is $0.5\mu A$ , and the range is 5 Amperes so, the current passing through the shunt resistance is $\left( {5 - 0.5 \times {{10}^{ - 6}}} \right) = 4.9999995A$
As we know, the potential drop across the resistances connected in parallel is the same. SO,
$
{V_g} = {V_{sh}} \\
\left( {0.5\mu A} \right)\left( {50} \right) = \left( {4.9999995} \right)\left( {{R_{sh}}} \right) \\
{R_{sh}} = \dfrac{{50 \times 0.5 \times {{10}^{ - 6}}}}{{4.9999995}} \\
= 5\mu \Omega \\
$
Hence, the value of the shunt resistance that should be connected in parallel with the ammeter is 5 micro-ohms.
Note:It is interesting to note here that as we need to increase the range of the ammeter so we have connected a very low value of the resistance so that a major part of the current inline will flow through it without harming the meter.
Recently Updated Pages
Class 12 Question and Answer - Your Ultimate Solutions Guide
Master Class 12 Social Science: Engaging Questions & Answers for Success
Master Class 12 Physics: Engaging Questions & Answers for Success
Master Class 12 Maths: Engaging Questions & Answers for Success
Master Class 12 English: Engaging Questions & Answers for Success
Master Class 12 Chemistry: Engaging Questions & Answers for Success
Trending doubts
Explain sex determination in humans with the help of class 12 biology CBSE
Give 10 examples of unisexual and bisexual flowers
How do you convert from joules to electron volts class 12 physics CBSE
Differentiate between internal fertilization and external class 12 biology CBSE
On what factors does the internal resistance of a cell class 12 physics CBSE
A 24 volt battery of internal resistance 4 ohm is connected class 12 physics CBSE