Question

When a voltage of 20 volt is applied between the two ends of a coil, 800 cals heat is produced in 1 second. The value of resistance of the coil is (1 calorie = 4.2 joule)

• A. 1.2 $\Omega$
• B. 1.4 $\Omega$
• C. 0.12 $\Omega$
• D. 0.14 $\Omega$

Answer 1

The correct option is C 0.12 $\Omega$

Energy ($power\times time$) is measured in Joules and by including time (t) in the power formulae, the energy dissipated by a component or circuit can be calculated.
Energy dissipated = Pt or $VI\times tasP=VI$.
It can also be written as $\frac{V^2}{R}\times tasI=\frac{V}{R}$.
Hence, the expression for Power in terms of voltage is given as $P=\frac{V^{2}t}{R}Joules$.
It is given that voltage is 200 V and 880 cal/s of heat is generated. That is 800*4.2 J = 3360 J/s.
Therefore, $3360J=\frac{{20}^2}{R}$. So, $R=\frac{{20}^2}{3360}=0.12\Omega$.
Hence, the resistance the electrical appliance is $0.12\Omega$.