In Fig.6.66, voltmeter is not ideal. If the voltmeter is removed from `R_(1)` and then put across `R_(2)`, what will be the effect on current `I`? Given
`R_(1) gt R_(2)`.

Options

A. decreases
B. remains same
C. increases
D. `I` would have been same if voltmeters were ideal.

Answer

A, D

Explanation

Let resistance of voltmeter be `R_(V)`. Then the equivalent resistance of circuit is
`R_(eq) = R_(2) + (R_(1) R_(V))/(R_(1) + R_(V)) = (R_(1)R_(2) + R_(2)R_(V) + R_(V)R_(1))/(R_(1) + R_(V)`
`I = (E)/(R_(eq)) = (E(R_(1) + R_(V)))/(R_(1)R_(2) + R_(2)R_(V) + R_(V) R_(1))`
On putting voltmeter across `R_(2)`, we can get current
`I' = (E(R_(1) + R_(V)))/(R_(1)R_(2) + R_(2)R_(V) + R_(V) R_(1))` as `R_(1) gt R_(2) rArr I gt I'`
If voltmeter is ideal, then in both cases `R_(1)` and `R_(2)` will be in series and current in both ceses will be `I = E//(R_(1) + R_(2))`.