Current Electricity 4 Question 1
1. In a meter bridge experiment, the circuit diagram and the corresponding observation table are shown in figure
(2019 Main, 10 April I)
S. No. | $R(\Omega)$ | $l(cm)$ |
---|---|---|
1. | 1000 | 60 |
2. | 100 | 13 |
3. | 10 | 1.5 |
4. | 1 | 1.0 |
Which of the readings is inconsistent?
(a) 3
(b) 2
(c) 1
(d) 4
(a) the current $I$ through the battery is $7.5 mA$
(b) the potential difference across $R _L$ is $18 V$
(c) ratio of powers dissipated in $R _1$ and $R _2$ is 3
(d) if $R _1$ and $R _2$ are interchanged, magnitude of the power dissipated in $R _L$ will decrease by a factor of 9
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Solution:
- Unknown resistance ’ $X$ ’ in meter bridge experiment is given by
$$ X=\frac{100-l}{l} R $$
Case (1) When $R=1000 \Omega$ and $l=60 cm$, then
$$ X=\frac{(100-60)}{60} \times 1000=\frac{40 \times 1000}{60} $$
$$ \Rightarrow \quad X=\frac{2000}{3} \Omega \approx 667 \Omega $$
Case (2) When $R=100 \Omega$ and $l=13 cm$, then
$$ \begin{aligned} X=\frac{100-13}{13} & \times 100 \\ & =\frac{100 \times 87}{13}=\frac{8700}{13} \Omega \approx 669 \Omega \end{aligned} $$
Case (3) When $R=10 \Omega$ and $l=1.5 cm$, then
$$ \begin{aligned} X & =\frac{100-1.5}{1.5} \times 10 \\ & =\frac{98.5}{1.5} \times 10=\frac{9850}{15} \Omega \approx 656 \Omega \end{aligned} $$
Case (4) When $R=1 \Omega$ and $l=1.0 cm$, then
$$ \begin{aligned} & & X=\frac{100-1}{1} \times 1 \\ \therefore & X & =99 \Omega \end{aligned} $$
Thus, from the above cases, it can be concluded that, value calculated in case (4) is inconsistent.