Current Electricity Ques 77
- A heater is designed to operate with a power of $1000 W$ in a $100 V$ line. It is connected in combination with a resistance of $10 \Omega$ and a resistance $R$, to a $100 V$ mains as shown in the figure. What will be the value of $R$ so that the heater operates with a power of $62.5 W$ ?
(1978)
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Solution:
- For balanced meter bridge experiment
$$ \begin{array}{rlrl} & & \frac{X}{R} & =\frac{l}{(100-l)} \quad(\text { where }, R=90 \Omega) \\ \therefore & & \frac{X}{90} & =\frac{40}{100-40} \\ \therefore & X & =60 \Omega \end{array} $$
$$ \begin{alignedat} X & =R \frac{l}{(100-l)} \\ \frac{\Delta X}{X} & =\frac{\Delta l}{l}+\frac{\Delta l}{100-l}=\frac{0.1}{40}+\frac{0.1}{60} \\ \Delta X & =0.25 \\ \text { So, } \quad X & =(60 \pm 0.25) \Omega \end{aligned} $$
BETA
