Current Electricity 4 Question 32
32. 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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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
$$ =V _{a b}=i^{\prime} R=\left(5.53 \times 10^{-3}\right)(285.71)=1.58 V $$
