• # question_answer A heater coil connected to a supply of a 220 V is dissipating some power ${{P}_{1}}.$ The coil is cut into half and the two halves are connected in parallel. The heater now dissipates a power ${{P}_{2}}.$ The ratio of power ${{P}_{1}}\,\,:\,\,{{P}_{2}}$ is         [AFMC 2004] A)            2 : 1                                           B)            1 : 2 C)            1 : 4                                           D)            4 : 1

$P=\frac{{{V}^{2}}}{R}$. If resistance of heater coil is R, then resistance of parallel combination of two halves will be $\frac{R}{4}$ So  $\frac{{{P}_{1}}}{{{P}_{2}}}=\frac{{{R}_{2}}}{{{R}_{1}}}=\frac{R/4}{R}=\frac{1}{4}$