A voltmeter has a 25Ω coil and 575Ω in series. The coil takes 10mA for full scale deflection. The maximum potential difference which can be measured is
A
250mV
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B
5.75V
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C
5.5V
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D
6.0V
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Solution
The correct option is D6.0V Let, resistance of voltmeter is, Rv=25Ω resistance of coil is, Rc=575Ω Current is I=10mA=10×10−3A Since, a voltmeter and a coil are connected in series, same current flow through both of them. According to KVL. IRv+IRc−V=0 V=I(Rv+Rc) V=10×10−3(25+575) V=600×10−2 V=6V