In the circuit shown in the figure- the potential difference across the 4-5 mu F capacitor is-displaystyle frac-8-3- volt4 volt6 volt8 volt

Question

Toppr Admin · Accepted Answer

Total capacitance- 1C-14-5-19-13or C-3-x3BC-FCharge- Q - CV - 3 -x3BC- F -xD7- 12V - 36 -x3BC- CPotential difference across 4-5-x3BC-F capacitor-36-x3BC-C4-5-x3BC-F-8V

In the circuit shown in the figure, the potential difference across the $4.5 μ F$ capacitor is:

$\frac{8}{3}$ volt
4 volt
6 volt
8 volt

Total capacitance, $\frac{1}{C} = \frac{1}{4.5} + \frac{1}{9} = \frac{1}{3}$
or $C = 3 μ F$
Charge, Q = CV = 3 $μ$ F $\times$ 12V = 36 $μ$ C
Potential difference across $4.5 μ F$ capacitor

$= \frac{36 μ C}{4.5 μ F} = 8 V$

In the circuit shown in the figure, the potential difference across the 4.5μF capacitor is:83 volt4 volt6 volt8 volt

Total capacitance, 1C=14.5+19=13or C=3μFCharge, Q = CV = 3 μ F × 12V = 36 μ CPotential difference across 4.5μF capacitor=36μC4.5μF=8V

In the circuit shown in the figure, the potential difference across the $4.5 μ F$ capacitor is:

$\frac{8}{3}$ volt
4 volt
6 volt
8 volt

Total capacitance, $\frac{1}{C} = \frac{1}{4.5} + \frac{1}{9} = \frac{1}{3}$
or $C = 3 μ F$
Charge, Q = CV = 3 $μ$ F $\times$ 12V = 36 $μ$ C
Potential difference across $4.5 μ F$ capacitor

$= \frac{36 μ C}{4.5 μ F} = 8 V$