Question

Can I please get help with these six questions? Thank you so much!

In the circuit below, the switch is closed after it had been open a long time. If the emf, resistances, and capacitance are e = 24 V, R1 = 6.0 2, R2 = 2.0 2, and C = 10.0 uf, what is the charge stored on the capacitor a long time after the switch is closed? R w to R:

Answer 1

1) In an RC dc circuit, after a very long time, a capacitor offers infinite resistance.

So, after a very long time, the resistors R1 and R2 are in series and the potential difference across the capacitor is equal to the potential difference across the resistor R2.

Total resistance after very long time is R = R1 + R2 = 8 ohm.

Current in the circuit = i = E/R = 24/8 = 3A

potential difference across R2 = V' = i*R2 = 3*2 = 6V

Therfore, charge stored in the capacitor = Q = C*V' = 10*6 = 60 micro coulomb. [answer]

3) The resistors are in series, therefore, net resistance is R = 9 + 3 = 12 ohm

current in the circuit = i = 2A

Therefore, battery voltage is V = iR = 2*12 = 24 V [answer]

4) The equation for charge on capacitor in RC circuit, when it decaying is $q = Q_{o}e^{-\frac{t}{RC}}$ .

At half-life, q = Qo/2.

therefore, half-life = $t_{1/2}=RC*ln2$

$\Rightarrow R = \frac{t_{1/2}}{C*ln2} = \frac{12*10^{-3}}{0.01*ln2}=1.73\Omega$[answer]

5) The equivalent resistance is $R_{eq.}=\left (\frac{1}{R}+\frac{1}{R}+\frac{1}{R} \right )^{-1}=\frac{R}{3}$

The equivalent capacitance is $C_{eq.}=\left (\frac{1}{C}+\frac{1}{C} \right )^{-1}=\frac{C}{2}$

The time constant is $\tau = R_{eq.}C_{eq.}=\frac{R}{3}*\frac{C}{2}=\frac{3*1}{6}=0.5s$ [answer]

6) Resistance = $R=\frac{\rho L}{A}$

Resistance of wire with circular cross-section = $R_{circ.}=\frac{\rho L_{circ.}}{A_{circ.}}$

$\Rightarrow 4.8*10^{-3}=\frac{\rho *2}{\pi*(1.5*10^{-3})^{2}}$---------(i)

Reistance of wire with square cross-section = $R_{sq.}=\frac{\rho L_{sq.}}{A_{sq.}}$

$\Rightarrow R_{sq.}=\frac{\rho*2}{(1.5*10^{-3})^{2}}$------(ii)

Dividing the equation (ii) by (i):

$\frac{R_{sq.}}{4.8*10^{-3}}=\frac{\frac{\rho*2}{(1.5*10^{-3})^{2}}}{\frac{\rho *2}{\pi*(1.5*10^{-3})^{2}}}$

$\Rightarrow R_{sq.}=\pi*4.8*10^{-3}=15.08*10^{-3}\Omega = 15.08m\Omega$ [answer]