A galvanometre of 100 ohm resistence coil shows 50 micro ampere whole difflex .how much resistance should be added to make (1)50 volt voltmetre(2)10 m a ametre
Answers
Answer:
The stored energy in the coil is 125 joules.
Given:
Self-inductance = 10 H = L
Resistance = 10 Ω = R
Voltage = 50 V = V
To find:
Energy stored in the coil = ?
Solution:
The energy stored in the magnetic field is given by the formula:
E=\frac{1}{2} L I_{0}^{2}E=
2
1
LI
0
2
Where,
L = Inductance
I = Current
Since, the value of current is not given.
On applying, Ohm's law, we can determine the value of current flowing through the coil. Thus, the current is given by the formula.
I_{0}=\frac{V}{R}I
0
=
R
V
On substituting the given values, we get,
I_{0}=\frac{50}{10} = 5 \ AI
0
=
10
50
=5 A
Now, on substituting values on the energy formula, we get,
E=0.5\times 10\times (5)^2E=0.5×10×(5)
2
\therefore E = 125 \ J∴E=125 J
Answer:
Explanation:in the case of increasing the range of current shunt is added to parallel to the galvanometer ok so to increase efficiency of increasing the range of current the shunt should be added parallel and in the case of increasing the range of voltage it means a conversion of galvanometer to voltmeter the shunt should be added in the series so in the case of making it into a ammeter the voltage will remain constant across the galvanometer and shunt and in the case of making it into a voltmeter shunt should be added in series it means the current will remain constant in galvanometer and shunt