In the adjoining figure, four capacitors are shown with their respective capacities and the PD applied. The charge and the PD across the 4μ F capacitor will be

Question

In the adjoining figure, four capacitors are shown with their respective capacities and the PD applied. The charge and the PD across the 4μ F capacitor will be (A)  600 μ C; 150 V  (B)  300 μ C; 75 V  (C)  800 μ C; 200V  (D)  580 μ C; 145V

Tardigrade Admin · Accepted Answer

Total capacitance (1/C)=(1/20)+(1/8)+(1/12) ⇒ C=(120/31)μ F Total charge Q=CV=(120/31)× 300=1161μ C Charge through 4μ F condenser =(1161/2)=500μ C and potential difference across it =(580/4)=145 V

Q. In the adjoining figure, four capacitors are shown with their respective capacities and the PD applied. The charge and the PD across the $4 μ F$ capacitor will be

$600 μ C; 150 V$

$300 μ C; 75 V$

$800 μ C; 200 V$

$580 μ C; 145 V$

Q. In the adjoining figure, four capacitors are shown with their respective capacities and the PD applied. The charge and the PD across the 4μF capacitor will be

600μC;150V

300μC;75V

800μC;200V

580μC;145V

Total capacitance C1​=201​+81​+121​ ⇒ C=31120​μF Total charge Q=CV=31120​×300=1161μC Charge through 4μF condenser =21161​=500μC and potential difference across it =4580​=145V

Q. In the adjoining figure, four capacitors are shown with their respective capacities and the PD applied. The charge and the PD across the $4 μ F$ capacitor will be

$600 μ C; 150 V$

$300 μ C; 75 V$

$800 μ C; 200 V$

$580 μ C; 145 V$