An electric device,which heats water by immersing a resistance wire in the water,generates 50 cal of heat per second when an electric potential difference of 12 V is placed across its leads.What is the resistance of the heater wire? (Note: 1 cal = 4.186 J)
A)0.94 $\Omega$
B)0.81 $\Omega$
C)0.58 $\Omega$
D)0.69 $\Omega$
E)1.5$\Omega$

Question

Quizplus · Accepted Answer

We can use the formula: Power = (Electric potential difference)^2 / Resistance

Here, the power is given as 50 cal/s, which can be converted to Joules/s as

50 cal/s x 4.186 J/cal = 209.3 J/s

The electric potential difference is given as 12 V.

Substituting these values, we get:

209.3 J/s = 12^2 / R

Solving for R, we get:

R = 0.69 Ω

Therefore, the correct answer is D.

An Electric Device,which Heats Water by Immersing a Resistance Wire Ω\OmegaΩ

An Electric Device,which Heats Water by Immersing a Resistance Wire $\Omega$