Current Electricity Charging Discharging Of Capacitors Question 93
Question: The resistance of a heater coil is 110 ohm. A resistance R is connected in parallel with it and the combination is joined in series with a resistance of 11 ohm to a 220 volt main line. The heater operates with a power of 110 watt. The value of R in ohm is
[ISM Dhanbad 1994]
Options:
A) 12.22
B) 24.42
C) Negative
D) That the given values are not correct
Show Answer
Answer:
Correct Answer: A
Solution:
Power consumed by heater is 110 W so by using $ P=\frac{V^{2}}{R} $
$ 110=\frac{V^{2}}{110}\Rightarrow ,V=110,V. $ Also from figure $ i _{1}=\frac{110}{110}=1A $ and $ i=\frac{110}{11}=10,A. $ So $ i _{2}=10-1=9,A $ Applying Ohms law for resistance R, V = iR
$ \Rightarrow 110=9\times R $
Therefore $ R=12.22,\Omega $
BETA
