1.

A circuit involving five ideal cells, three resistors (R_(1), R_(2) and 20 Omega) and a capacitor of capacitance C=1 mu F is shown. At stead state match the proper entries from column -2 to column -1 using the codes given below the columns. {:(,"Column" I,,"Column" II,),(,(P)K_(2) "is open and" K_(2) "is in position" C,,(1)"Potential at point A is greater than potential at B",),(,(Q)K_(2)"is open and" K_(1) "is in position D",,(2)"Current through" R_(1) "is downward",),(,(R)K_(2)"is closed and" K_(1) "is in positionC",,(3)"Current through" R_(2) "is upward",),(,(S)K_(2) "is closed and" K_(1) "is in position"C,,(4)"Charge on capacitor is" 10muC.,):}

Answer»

`{:((P),(Q),(R),(S),),(4,3,1,2,):}`
`{:((P),(Q),(R),(S),),(3,1,4,2,):}`
`{:((P),(Q),(R),(S),),(3,2,1,4,):}`
`{:((P),(Q),(R),(S),),(1,4,3,2,):}`

Solution :The state of key `K_2` has no effect on current through `R_1 and R_2` as well has no effect on charge in the capacitor. Also position of key `K_1` has no effect on potential difference between POINTS A and B, that is `V_A-V_B=10` volts under all conditions. Hence charge on capacitor under all cases in `10muC`.
Assume the potential at point P to be zero.
When Key `K_1` is in position C: `V_A`=16 Volt and `V_B`=6 volts. Hence current in both `R_1 and R_2` will flow downwards.
When Key `K_1` is in position D : `V_A`=2 Volt and `V_B`=-8 volts. Hence current through `R_1` wil flow downwards and through `R_2` will flow upwards.


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