Question

In the adjoining circuit, the emf of the cell is $2$ volt and the internal resistance is negligible. The resistance of the voltmeter is $80ohm$. The reading of the voltmeter will be

• A. 0.80 volt
• B. 1.60 yolt
• C. 1.33 volt
• D. 2.00 volt

Answer 1

Answer: (C)

Total resistance of the circuit $=\frac{80}{2}+20=60\Omega$
$\Rightarrow$ Main current, $i=\frac{2}{60}=\frac{1}{30}A$
Combination of voltmeter and $80\Omega$ resistance is connected in series with $20\Omega$, so current through $20\Omega$ and this combination will be same, i.e., $\frac{1}{30}A$.
Since the resistance of voltmeter is also $80\Omega$, so this current is equally distributed in $80\Omega$ resistance and voltmeter (i.e., $\frac{1}{60}$ A through each)
Potential difference across. $80\Omega$ resistance $=\frac{1}{60}\times 80=1.33V$