Current Electricity Ques 96
- Two resistors, $400 $ $\Omega$ and $800 $ $\Omega$ are connected in series with a $6 $ $V$ battery. It is desired to measure the current in the circuit. An ammeter of $10 $ $\Omega$ resistance is used for this purpose. What will be the reading in the ammeter? Similarly, if a voltmeter of $1000 $ $\Omega$ resistance is used to measure the potential difference across the $400 $ $\Omega$ resistor, what will be the reading in the voltmeter?
$(1982,6 M)$
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Answer:
Correct Answer: 96.$(4.96$ $ mA$, $1.58$ $ V)$
Solution:
$1000 $ $\Omega$ (a)
(b)
Refer figure (a) Current through ammeter,
$ \begin{aligned} i & =\frac{\text { net emf }}{\text { net resistance }} \\ & =\frac{6}{400+800+10} \\ & =4.96 \times 10^{-3} A \\ & =4.96 mA \end{aligned} $
Refer figure (b) Combined resistance of $1000 $ $\Omega$ voltmeter and $400 $ $\Omega$ resistance is,
$ \begin{aligned} R & =\frac{1000 \times 400}{1000+400}=285.71 \Omega \\ \therefore \quad i & =\frac{6}{(285.71+800)}=5.53 \times 10^{-3} A \end{aligned} $
Reading of voltmeter
$ =V _{a b}=i^{\prime} R=\left(5.53 \times 10^{-3}\right)(285.71)=1.58 $ $V $
BETA
