Question

STATE
7) A galvanometer of resistance 50 Q gives a full
scale deflection for a current 5 x 10-A The
resistance tljat should be connected in series
with the galvanometer to read 3 V is.
A) 5050 0 B) 5950 o C) 595 0 D) 5059 0
S
GELENE
d
HET VERLER
Hinterest
UTIH
TIDA
EPERTI
PERU
TOUTE​

Answer 1

Correct Question

A galvanometer of resistance 50Ω gives a fullscale deflection for a current 5 x 10-¹mA The resistance that should be connected in series with the galvanometer to read 3 V is

A) 5050 Ω B) 5950 Ω C) 595 Ω D) 5059 Ω

Solution

Given that, the resistance of galvanometer is 50Ω and the full-scale deflection current is 5×10-¹ mA or 0.5 mA.

(0.5 mA = 0.5 × 10-³ A)

We have to find the resistance (R) that should be connected in series with the galvanometer to read 3 V.

Now,

V = IR

(Voltage = Current × Resistance)

R = V/I

R = 3/(0.5 × 10-³)

R = 30000/5 = 6000Ω

Given that, resistance is connected I. series. So,

Rs = R - Rg

(Resistance in series = Resistance - Resistance of galvanometer)

Rs = 6000 - 50 = 5950 Ω

Option B) 5950 Ω

Answer 2

Given:-

A galvanometer of resistance 50Ω gives a full scale deflection for a current $5* 10^-^1$ mA.

To find:-

The resistance that should be connected in series with the galvanometer to read 3 V.

Solution:-

We know that,

Voltage, V = Current, I * Resistance, R

or, V = IR

or, R = V/I

P.D. = 3 V

I = $5* 10^-^1$ mA = 0.5 mA = 0.5/1000 = 0.5/10³ = $0.5*10^-^3$ A

R = V/I

⇒R = 3/($0.5*10^-^3$)

⇒R = 3/0.0005

⇒R = 6000 Ω

As the resistance is in series, so the value is:

Resistance in series = Resistance - Resistance of galvanometer

⇒$\sf{R_s = 6000-50}$

⇒$\sf{R_s$ = (B)5950 Ω