1.

Consider the circuit given here. The potential difference V_(BC) between the points B and C is

Answer»

`1V`
`0.5 V`
`0V`
`-1 V`

Solution :
currentflowingin the circuit
`i=(V )/( R_1+R_2 +R_3 ) = (3)/( (1 + 2+3 ) xx 10^3 ) = (3)/( 6 xx 10^3 )`
`impliesi= 0.5xx 10^(-3)` A
Potentialdropacrossthe armAD
`V_(AD )= IR = 0.5xx 10^(-3)xx 3 xx 10^(-3) = 1.5 V `
thechargeis GIVENBY` Q=CV= ((1 xx 2 )/( 2 +1)) xx 1.5 `
`= 2/3xx 1.5= 1 mu C `
potentialatpointB ` V_B= V_(AD) -V_(AQ) `
` = 1.5V -( 00.5m Axx 1 K Omega) =1 V `
Potentialatpoint` C , V_C= V_(AD) -V_(AP)`
`=1.5V -(1 mu C )/( 1 mu F) = 1.5V- = 0.5V `
` thereforeV_B-V_C- 1V- 0.5V = 0.5V `


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