Question

The coil of a measuring instrument has a resistance of 1 Ω and the instrument has a full scale deflection of 250 V when a resistance of 4999 Ω is connected with it. Find the current range of the instrument when used as an ammeter with the coil connected across a shunt of (1/499) Ω and the value of the shunt resistance for the instrument to give a full scale deflection of 50 A.​

Answer 1

V=IR

therefore current in the series circuit I=

5000+2

250

=0.04998A

current through shunt is I

sh

=

R

sh

I

m

R

m

=

2×10

−

3

0.04998×2

=49.98A

current range of instrument I

m

+I

sh

=49.98+0.04998=50A

Answer 2

Answer:

The current range of instruments is 25.05 A,

Shunt Resistance = 5.005 ohm

Step-by-step explanation:

Given:

The coil of a measuring instrument has a resistance of 1 Ω and the instrument has a full-scale deflection of 250 V when a resistance of 4999 Ω is connected with it.

R = 4999 + 1 ohm

V = 250

To find:

The current range of the instrument when used as an ammeter with the coil connected across a shunt of (1/499) Ω and the value of the shunt resistance for the instrument to give a full-scale deflection of 50 A.​

Using ohms law, V = IR.

Therefore current in the series circuit I = $\frac{250}{4999+1}$ = 0.05A

Current through the shunt is $I _{sh}$ = $\frac{I _m R _m}{ R _{sh}}$,

                                                     = $\frac{0.05 \times 1}{{1}{499}}$,

                                                      = $\frac{0.05 \times 1}{0.002}$,

                                                      = 25 A

The current range of instruments $I _m + I _{sh}$ = 0.05 + 25 = 25.05 A

$R_m = 4999 + 1$ =5000 ohm

Full scale deflection current = 50A

$R_{sh} = \frac{R_m}{[\frac{I}{I_m}-1]}$,

      = $\frac{5000}{[\frac{50}{0.05}-1]}$,

     = $\frac{5000}{999}$,

     = 5.005 ohm

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