Question

Two moving coil meters M1 and M2 have the following particular $\mathrm { R } _ { 2 } = 14 \Omega ; \mathrm { N } _ { 2 } = 42 ; \mathrm { A } _ { 2 } = 1.8 \times 10 ^ { - 3 } \mathrm {~m} ^ { 2 } ; \mathrm { B } _ { 2 } = 0.50 \mathrm {~T}$ The spring constants are indentical for the two metres. What is the ratio of corrent sensitivety and voltate sensitivity of M2 to M1?

• A. 1.4, 1
• B. 1.4, 0
• C. 2.8, 2
• D. 2.8, 0

Answer 1

Answer: (A)

1.4, 1

Answer 2

For meter $\mathrm { M } _ { 1 } , \mathrm { R } _ { 1 } = 10 \Omega ; \mathrm { N } _ { 1 } = 30$

For meter $\mathrm { M } _ { 2 } , \mathrm { R } _ { 2 } = 14 \Omega ; \mathrm { N } _ { 2 } = 42$

$\mathrm { A } _ { 2 } = 1.8 \times 10 ^ { - 3 } \mathrm {~m} ^ { 2 } ; \mathrm { B } _ { 2 } = 0.50 \mathrm {~T} ; \mathrm { K } _ { 2 } = \mathrm { k }$

So, current sensitivity

and voltage sensitivity $V _ { s } = \frac { N _ { 2 } B A } { k R }$

Now,

$\frac { V _ { s _ { 2 } } } { V _ { s _ { 1 } } } = \frac { N _ { 2 } B _ { 2 } A _ { 2 } / \left( K _ { 2 } R _ { 2 } \right) } { N _ { 1 } B _ { 1 } A _ { 1 } / \left( K _ { 1 } R _ { 1 } \right) } = \frac { N _ { 2 } B _ { 2 } A _ { 2 } R _ { 1 } K _ { 1 } } { N _ { 1 } B _ { 1 } A _ { 1 } R _ { 2 } K _ { 2 } }$

$= \frac { 42 \times 0.50 \times \left( 1.8 \times 10 ^ { - 3 } \right) \times 10 \times \mathrm { k } } { 30 \times 0.25 \times \left( 3.6 \times 10 ^ { - 3 } \right) \times 14 \times \mathrm { k } } = 1$