A $100\, watt$ bulb working on $200 \,volt$ and a $200\, watt$ bulb working on $100\, volt$ have
AResistances in the ratio of $4 : 1$
BMaximum current ratings in the ratio of $1:4$
CResistances in the ratio of $2 : 1$
DMaximum current ratings in the ratio of $1 : 2$
Medium
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BMaximum current ratings in the ratio of $1:4$
b (b) $R = \frac{{{V^2}}}{P}$ $ \Rightarrow $ ${R_1} = \frac{{200 \times 200}}{{100}} = 400\,\Omega $ and
${R_2} = \frac{{100 \times 100}}{{200}} = 50\,\Omega .$ Maximum current rating $i = \frac{P}{V}$
So ${i_1} = \frac{{100}}{{200}}$ and ${i_2} = \frac{{200}}{{100}}$$ \Rightarrow $ $\frac{{{i_1}}}{{{i_2}}} = \frac{1}{4}$.
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An electric bulb rated as $200\, {W}$ at $100 \,{V}$ is used in a circuit having $200\, {V}$ supply. The resistance $'R'$ that must be put in series with the bulb so that the bulb delivers the same power is $.....\,\Omega$
In the circuit shown in the figure $K_1$ is open. The charge on capacitor $C$ in steady state is $q_1$. Now key is closed and at steady state charge on $C$ is $q_2$. The ratio of charges $q_1/q_2$ is
Following figures show different combinations of identical bulb$(s)$ connected to identical battery$(ies)$. Which option is correct regarding the total power dissipated in the circuit?