Question

Potentiometer wire of length 1 m is connected in series with 490Ω resistance and 2 V battery. If 0.2 mV/cm is the potential gradient, then resistance of the potentiometer wire is(a) 4.9 W(b) 7.9 W(c) 5.9 W(d) 6.9 W

Answer 1

Answer:

A) 4.9 W

Explanation:

Length of wire = L = 1m

Resistance = Rh = 490Ω

Battery = e = 2v

i = [e / (R + Rh + r)]

 i = {2 / (R + 490 + 0)} = {2 / (R + 490)}

Potential gradient = (v/L) = (iR / L) = [2 / (R + 490)] × R × (1/1)

Potential gradient = 2R / (R + 490)

[(0.2 × 10–3) / (10–2)] (v/m) = {2R / (R + 490)}

Thus,

0.02(R + 490) = 2R

0.01 = R / (R + 490)

0.01 R + 4.9 = R

0.99 R = 4.9

R = 4.94Ω