Current Electricity Charging Discharging Of Capacitors Question 254
Question: In the adjoining circuit, the e.m.f. of the cell is 2 volt and the internal resistance is negligible. The resistance of the voltmeter is 80 ohm. The reading of the voltmeter will be
[CPMT 1991]
Options:
A) 0.80 volt
B) 1.60 volt
C) 1.33 volt
D) 2.00 volt
Show Answer
Answer:
Correct Answer: C
Solution:
Total resistance of the circuit $ =\frac{80}{2}+20=60,\Omega $
Therefore Main current $ i=\frac{2}{60}=\frac{1}{30}A $ Combination of voltmeter and 80W resistance is connected in series with 20W, so current through 20W and this combination will be same $ =\frac{1}{30}A $ . Since the resistance of voltmeter is also 80W, so this current is equally distributed in 80W resistance and voltmeter (i.e. $ \frac{1}{60}A $ through each) P.D. across 80W resistance $ =\frac{1}{60}\times 80=1.33,V $
BETA
