In the arrangement shown initially switch s is open. When switch s is closed,
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a
reading of voltmeter is halned
b
reading of voltmeter becomes 4 times the initial reading
c
power dissipated in left resistor becomes 4 times the initial power
d
power dissipated in the left resister is doubled
answer is C.
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Detailed Solution
When s is open, i = E/2R∴ reading of voltmeter = i R= E/2 .Initial power dissipated in the left resistor = i2R=E2R2R=E2/4R.When s is closed, reading of voltmeter = E. Power dissipated in left resistor = E2R.