A galvanometer having a resistance of 20 Ω and 30 division on both sides has figure of merit 0.005 ampere/division. The resistance (in ohms) that should be connected in series such that it can be used as a voltmeter upto 15 volt, is:

Question

A galvanometer having a resistance of 20 Ω and 30 division on both sides has figure of merit 0.005 ampere/division. The resistance (in ohms) that should be connected in series such that it can be used as a voltmeter upto 15 volt, is: (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Deflection current Igmax = n × k = 0.005 × 30 Where, n = Number of divisions = 30 and k = 0.005 amp/division =15× 10-2 =0.15 v=Ig[20+R] 15=0.15 [20+R] 100=20+R R=80 Ω

Q. A galvanometer having a resistance of $20 Ω$ and $30$ division on both sides has figure of merit $0.005$ ampere/division. The resistance (in ohms) that should be connected in series such that it can be used as a voltmeter upto $15$ volt, is:

Deflection current
$I g_{ma x} = n \times k = 0.005 \times 30$
Where, n = Number of divisions = 30 and k = 0.005 amp/division
$= 15 \times 1 0^{- 2} = 0.15$
$v = I_{g} [20 + R]$
$15 = 0.15 [20 + R]$
$100 = 20 + R$
$R = 80 Ω$

Q. A galvanometer having a resistance of 20Ω and 30 division on both sides has figure of merit 0.005 ampere/division. The resistance (in ohms) that should be connected in series such that it can be used as a voltmeter upto 15 volt, is:

Deflection current Igmax​=n×k=0.005×30 Where, n = Number of divisions = 30 and k = 0.005 amp/division =15×10−2=0.15 v=Ig​[20+R] 15=0.15[20+R] 100=20+R R=80Ω

Q. A galvanometer having a resistance of $20 Ω$ and $30$ division on both sides has figure of merit $0.005$ ampere/division. The resistance (in ohms) that should be connected in series such that it can be used as a voltmeter upto $15$ volt, is:

Deflection current
$I g_{ma x} = n \times k = 0.005 \times 30$
Where, n = Number of divisions = 30 and k = 0.005 amp/division
$= 15 \times 1 0^{- 2} = 0.15$
$v = I_{g} [20 + R]$
$15 = 0.15 [20 + R]$
$100 = 20 + R$
$R = 80 Ω$