Three capacitors of capacitance 2 μ F , 3 μ F and 6 μ F are connected in series and the combination is charged by means of a 24 volt battery. The potential difference between the plates of the 6 μ F capacitor will be :

Question

Three capacitors of capacitance 2 μ F , 3 μ F and 6 μ F are connected in series and the combination is charged by means of a 24 volt battery. The potential difference between the plates of the 6 μ F capacitor will be : (A) 10 V (B) 8 V (C) 4 V (D) 6 V

Tardigrade Admin · Accepted Answer

(1/Ce q)=(1/2)+(1/3)+(1/6) ⇒ Ce q=1 μ F Total charge Q=Ce q ⋅ V=1 × 24=24 μ C Potential diff. across 6 μ F capacitor =(24/6)=4 volt

Q. Three capacitors of capacitance $2 μ F, 3 μ F$ and $6 μ F$ are connected in series and the combination is charged by means of a $24$ volt battery. The potential difference between the plates of the $6 μ F$ capacitor will be :

10 V

8 V

4 V

6 V

$\frac{1}{C _{e q}} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6}$
$\Rightarrow C_{e q} = 1 μ F$
Total charge $Q = C_{e q} \cdot V = 1 \times 24 = 24 μ C$
Potential diff. across $6 μ F$ capacitor $= \frac{24}{6} = 4$ volt

Q. Three capacitors of capacitance 2μF,3μF and 6μF are connected in series and the combination is charged by means of a 24 volt battery. The potential difference between the plates of the 6μF capacitor will be :

10 V

8 V

4 V

6 V

Ceq​1​=21​+31​+61​ ⇒Ceq​=1μF Total charge Q=Ceq​⋅V=1×24=24μC Potential diff. across 6μF capacitor =624​=4 volt

Q. Three capacitors of capacitance $2 μ F, 3 μ F$ and $6 μ F$ are connected in series and the combination is charged by means of a $24$ volt battery. The potential difference between the plates of the $6 μ F$ capacitor will be :

$\frac{1}{C _{e q}} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6}$
$\Rightarrow C_{e q} = 1 μ F$
Total charge $Q = C_{e q} \cdot V = 1 \times 24 = 24 μ C$
Potential diff. across $6 μ F$ capacitor $= \frac{24}{6} = 4$ volt