Question

1. A capacitor of capacitance 2 uF is connected as shown in the
figure the internal resistance of battery is 0.5 ohm the amount of charge on capacitor in steady state is






(2) 2 uc
(1) Zero
(3) 4 uc
(4) 6 uc​

Answer 1

Answer:

The charge is 4 μF.

(3) is correct option.

Explanation:

Given that,

Capacitance = 2μF

Internal resistance = 0.5

We need to calculate the net resistance

Using parallel formula of resistance

$\dfrac{1}{R}=\dfrac{1}{R_{1}}+\dfrac{1}{R_{2}}$

Put the value into the formula

$\dfrac{1}{R}=\dfrac{1}{2}+\dfrac{1}{10}$

$R=\dfrac{5}{3}\ ohm$

Using series formula of resistance

$R'=R+R_{int}$

Put the value into the formula

$R'=\dfrac{5}{3}+0.5$

$R'=\dfrac{13}{6}\ ohm$

$R'=2.1\ ohm$

We need to calculate the current

Using Ohm's law

$V= IR$

$I =\dfrac{V}{R}$

Put the value into the formula

$I=\dfrac{2.5\times6}{13}$

$I=1.2\ A$

We need to calculate the potential drop across two parallel branches

Using formula of potential

$V=E-ir$

Put the value into the formula

$V=2.5-1.2\times0.5$

$V=1.9\ V$

We need to calculate the charge

Using formula of charge

$q=cV$

Put the value into the formula

$q=2\times10^{-6}\times1.9$

$q=3.8\ \mu F\approx4\ \mu F$

Hence, The charge is 4 μF.