The resistance of a heater coil is 110, Omega | vedclass

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(A) $1$. The power consumed by the heater is $P = 110\, W$ and its resistance is $R_h = 110\, \Omega$. Using the formula $P = \frac{V_h^2}{R_h}$,we fi

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$12.22$

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(A) $1$. The power consumed by the heater is $P = 110\, W$ and its resistance is $R_h = 110\, \Omega$. Using the formula $P = \frac{V_h^2}{R_h}$,we find the voltage across the heater:$110 = \frac{V_h^2}{110} \Rightarrow V_h^2 = 12100 \Rightarrow V_h = 110\, V$.$2$. The current through the heater is $i_1 = \frac{V_h}{R_h} = \frac{110}{110} = 1\, A$.$3$. The total voltage is $220\, V$. Since the heater combination is in series with an $11\, \Omega$ resistor,the voltage across the $11\, \Omega$ resistor is $V_s = 220 - 110 = 110\, V$.$4$. The total current $i$ flowing through the $11\, \Omega$ resistor is $i = \frac{V_s}{11} = \frac{110}{11} = 10\, A$.$5$. By Kirchhoff's Current Law,the current through $R$ is $i_2 = i - i_1 = 10 - 1 = 9\, A$.$6$. Since $R$ is in parallel with the heater,the voltage across $R$ is also $110\, V$. Thus,$R = \frac{V_h}{i_2} = \frac{110}{9} \approx 12.22\, \Omega$.

The resistance of a heater coil is $110\, \Omega$. $A$ resistance $R$ is connected in parallel with it,and the combination is joined in series with a resistance of $11\, \Omega$ to a $220\, V$ main line. The heater operates with a power of $110\, W$. The value of $R$ in $\Omega$ is

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