Two capacitors C_1 and C_2 in a circuit are j | vedclass

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

Question

(C) The capacitors $C_1$ and $C_2$ are connected in series between points $A$ and $B$.Since they are in series,the charge $Q$ on both capacitors must

Vedclass · Accepted Answer

$\frac{C_1 V_1 + C_2 V_2}{C_1 + C_2}$

Vedclass · Answer

(C) The capacitors $C_1$ and $C_2$ are connected in series between points $A$ and $B$.Since they are in series,the charge $Q$ on both capacitors must be the same.Let $V_D$ be the potential at point $D$.The potential difference across $C_1$ is $(V_1 - V_D)$ and across $C_2$ is $(V_D - V_2)$.Since $Q = C_1(V_1 - V_D) = C_2(V_D - V_2)$,$C_1 V_1 - C_1 V_D = C_2 V_D - C_2 V_2$$C_1 V_1 + C_2 V_2 = V_D(C_1 + C_2)$$V_D = \frac{C_1 V_1 + C_2 V_2}{C_1 + C_2}$

Two capacitors $C_1$ and $C_2$ in a circuit are joined as shown in the figure. The potential of point $A$ is $V_1$ and that of $B$ is $V_2$. The potential of point $D$ will be

Similar Questions

If the rate of the reaction is equal to the rate constant,the order of the reaction is

If $R$ is the set of all real numbers and $f: R-\{2\} \rightarrow R$ is defined by $f(x)=\frac{2+x}{2-x}$ for $x \in R-\{2\}$,find the range of $f(x)$.

The work function of a metallic surface is $5.01 \ eV$. The photo-electrons are emitted when light of wavelength $2000 \ \mathring{A}$ falls on it. The potential difference applied to stop the fastest photo-electrons is .............. $volts$ $[h = 4.14 \times 10^{-15} \ eV \cdot s]$

The shortest distance between the line $y - x = 1$ and the curve $x = y^2$ is:

If $g(x) = \int_0^x \cos^4 t \, dt$,then $g(x + \pi)$ equals

Vedclass Test Series

Exam Paper Generator

Online Exam Module