Question

Three capacitors of capacitances 1µf, 2µF & 3µF are connected in series and a potential difference of 11V is applied across the combination then the potential difference across the plates of 1µF capacitor is

Answer 1

Answer

Potential difference across the plates of 1µF capacitor is 6 V

Given

Three capacitors of capacitance's 1µF, 2µF & 3µF are connected in series and a potential difference of 11 V is applied across the combination

To Find

Potential difference across  the plates of 1µF capacitor

Formula Used

$\bigstar \;\; \bf Q=CV$

where ,

Q denotes charge

C denotes capacitance

V denotes potential difference

Solution

Find attachment

We know that , when potential difference is applied to the circuit . It is divided into the the following ,

$\implies \rm V_1+V_2+V_3=V\\\\\implies \rm V_1+V_2+V_3=11\ ...(1) \; [Given]$

For 1st , 2nd and 3rd capacitance , charge is flowing as follows ,

$\implies \rm Q=(C_1V_1)=(C_2V_2)=(C_3V_3)\\\\\implies \rm Q=((1)V_1)=((2)V_2)=((3)V_3)\\\\\implies \rm Q=V_1=2V_2=3V_3\\\\\rm V_2=\dfrac{V_1}{2},V_3=\dfrac{V_1}{3}$

Sub. these values in (1) , we get ,

$\implies \rm V_1+\dfrac{V_1}{2}+\dfrac{V_1}{3}=11\\\\\implies \rm \dfrac{6v_1+3V_1+2V_1}{6}=11\\\\\implies \rm 11V_1=66\\\\\implies \rm V_1=\dfrac{66}{11}\\\\\implies \rm V_1=6\ V$

So , Potential difference across 1µF capacitor is , V₁ = 6 V