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Question

Take the potential of the point B in the figure, to be zero. (a) Find the potentials at the points C and D. (b) If a capacitor is connected between C and D, What charge will appear on the capacitor?


1022360_490727152a3d4306920b25ed3dec926b.png


Solution

(a) Capacitance on upper branch = $$\dfrac { (4\times 8) }{ (4+8) } =\dfrac { 8 }{ 3 } \mu F$$
and capacitance on the lower branch= $$\dfrac { 6\times 3 }{ 6+3 } =\quad 2\mu F$$
(I) The charge on the capacitor $$\dfrac { 8 }{ 3 } \mu F$$ is Q=CV= $$\dfrac { 8 }{ 3 } \times 50=\frac { 400 }{ 3 } C$$
Therefore the potential difference at $$4\mu F$$=$$\dfrac { 400 }{ 3\times 4 } =\dfrac { 100 }{ 3 } $$
And at
 $$8\mu F$$ = $$\dfrac { 400 }{ 3\times 8 } =\dfrac { 100 }{ 6 } $$
 The potential difference= $$\dfrac { 100 }{ 3 } -\dfrac { 100 }{ 6 } =50\mu V$$

(ii) Effective change  at 2
$$\mu F= 50\times 2=100\mu F$$
Therefore potential difference at $$3\mu F=\dfrac { 100 }{ 3 } ,$$
Potential at 6$$\mu F=\dfrac { 100 }{ 6 } $$
Therefore difference=$$\dfrac { 100 }{ 3 } -\dfrac { 100 }{ 6 } =\dfrac { 50 }{ 3 } \mu V$$

Therefore pot.diff at C and D=$$\dfrac { 50 }{ 3 } \mu V$$

Physics

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