JEE Main & Advanced Physics Current Electricity, Charging & Discharging Of Capacitors / वर्तमान बिजली, चार्ज और कैपेसिटर का निर JEE PYQ-Current Electricity Charging and Discharging of Capacitors

A wire when connected to 220 V mains supply has power dissipation \[{{P}_{1}}\]. Now, the wire is cut into two equal pieces which are connected in parallel to the same supply. Power dissipation in this case is \[{{P}_{2}}\]. Then, \[{{P}_{2}}:{{P}_{1}}\] is [AIEEE 2002]

A) 1

B) 4

C) 2

D) 3

Correct Answer: B

Solution :

[b] In 1st case

Using the formula,\[P=\frac{{{V}^{2}}}{R}\] ... (i)

where, R is resistance of wire, V is voltage across wire and P is power dissipation in wire and

\[R=\frac{\rho l}{A}\] ... (ii)

From Eqs. (i) and-(ii), we get

\[{{P}_{1}}=\frac{{{V}^{2}}}{\rho l/A}=\frac{{{V}^{2}}}{\rho l}.\,A\]

\[{{P}_{1}}=\frac{{{V}^{2}}}{\rho l}.\,A\] ... (iii)

In 2nd case

Let \[{{R}_{2}}\] be net resistance.

\[{{R}_{2}}=\frac{R\times R}{R+R}=\frac{R}{2}\]

where, R is the resistance of half wire.

\[\therefore \] \[{{R}_{2}}=\frac{\rho .\left( \frac{l}{2} \right)}{A.\,2}=\frac{\rho l}{4A}\]

(as \[l=\frac{l}{2}\Rightarrow A=2A\])

\[\therefore \] \[{{P}_{2}}=\frac{{{V}^{2}}}{\rho l}.4\,A\] ... (iv)

Hence, from Eqs. (iii) and (iv), we get

\[\frac{{{P}_{1}}}{{{P}_{2}}}=\frac{1}{4}\Rightarrow \frac{{{P}_{2}}}{{{P}_{1}}}=\frac{4}{1}\]