In the following diagram,the charge and poten | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The given circuit can be simplified by identifying short-circuited components. The capacitors $9\, \mu F, 9\, \mu F$ and $7\, \mu F$ are connected

Vedclass · Accepted Answer

$214 \,\mu C, 27\, V$

Vedclass · Answer

(C) The given circuit can be simplified by identifying short-circuited components. The capacitors $9\, \mu F, 9\, \mu F$ and $7\, \mu F$ are connected in parallel with a wire,effectively short-circuiting them. Thus,they can be removed from the circuit.After simplification,the circuit consists of a parallel combination of $6\, \mu F, 4\, \mu F$ and $8\, \mu F$ capacitors,which is in series with a $36\, \mu F$ capacitor.The equivalent capacitance of the parallel part is $C_p = 6 + 4 + 8 = 18\, \mu F$.The total circuit is now a series combination of $C_p = 18\, \mu F$ and $C_s = 36\, \mu F$ connected to a $40\, V$ source.Let $V_1$ be the potential difference across the $18\, \mu F$ combination and $V_2$ be the potential difference across the $36\, \mu F$ capacitor.We have $V_1 + V_2 = 40\, V$ and the ratio of potential differences in series is inversely proportional to capacitance: $\frac{V_1}{V_2} = \frac{36}{18} = 2$,so $V_1 = 2V_2$.Substituting this into the sum: $2V_2 + V_2 = 40 \implies 3V_2 = 40 \implies V_2 = \frac{40}{3}\, V$.Then $V_1 = 40 - \frac{40}{3} = \frac{80}{3}\, V \approx 26.67\, V$.The potential difference across the $8\, \mu F$ capacitor is $V_1 = \frac{80}{3}\, V \approx 26.67\, V$.The charge on the $8\, \mu F$ capacitor is $Q = C 	imes V_1 = 8\, \mu F 	imes \frac{80}{3}\, V = \frac{640}{3}\, \mu C \approx 213.33\, \mu C$.Rounding to the nearest given option,the values are approximately $214\, \mu C$ and $27\, V$.

In the following diagram,the charge and potential difference across $8\, \mu F$ capacitance will be respectively

Similar Questions

The potential difference between the points $P$ and $Q$ in the adjoining circuit will be :-

Two capacitors of capacities $1 \mu F$ and $C \mu F$ are connected in series and the combination is charged to a potential difference of $120 \ V$. If the charge on the combination is $80 \mu C$,the energy stored in the capacitor of capacity $C$ in $\mu J$ is

In the circuit shown in the figure,four capacitors are connected to a battery. $A$ parallel plate capacitor has an electric field of $10^5 \ V/m$ between its plates. If the charge on the capacitor plate is $1 \ \mu C$,then the force on each capacitor plate is.....$N$.

$A$ charged capacitor is disconnected from the battery and if the distance between the two plates of the capacitor is increased then . . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module