JEE Advanced · 2014 · 25 May 2014 · Paper 1 · Current Electricity

A galvanometer gives full scale deflection with

0.006 A

current. By connecting it to a 4990

Ω

resistance, it can be converted into a voltmeter of range

0 - 30 V

. If connected to a

\frac{2 n}{249} Ω

resistance, it becomes an ammeter of range

0 - 1.5 A

. The value of

n

Answer: 5

Solution

Voltmeter:

I_{g} (R_{g} + R_{s}) = V

, so

0.006 (R_{g} + 4990) = 30

, giving

R_{g} + 4990 = 5000

, thus

R_{g} = 10

Ω.

Ammeter: Shunt resistance

R_{s h} = \frac{2 n}{249}

Ω. Current through galvanometer is

I_{g} = 0.006

A, through shunt is

I - I_{g} = 1.5 - 0.006 = 1.494

A.

Voltage across both:

I_{g} R_{g} = (I - I_{g}) R_{s h}

, so

0.006 \times 10 = 1.494 \times \frac{2 n}{249}

.

Solving:

0.06 = \frac{2.988 n}{249}

, so

n = \frac{0.06 \times 249}{2.988} = \frac{14.94}{2.988} = 5

JEE Advanced · 2014 · 25 May 2014 · Paper 1 · Integer answer