Question

In the circuit shown in the figure, the capacitor is initially uncharged and the key K is initially open in this condition, a current of 1 A flows through the 1 Ω resistor. The key is closed at a time t=t0. Choose the correct options(given $e^{-1}=0.36$)

• A. R=3Ω
• B. Before t=t0, i1-2A
• C. At t=t0+7.2μ sec. current in the capacitor =0.6A
• D. At t→∞, charge on the capacitor is 12μC

Answer 1

Answer: (A), (B), (C), (D)

R=3Ω

Answer 2

$\frac{x-5}{1}\Rightarrow\,\,x=+6\,,\,i_1=volt,\frac{6-0}{3}=2A$
R=$\frac{15-6}{3}$=3$\Omega$
after switching on :
$\Rightarrow$

$\varepsilon_{eq}=\frac{\frac{15}{3}+\frac{5}{1}+\frac{0}{3}}{\frac{1}{2}+\frac{1}{1}+\frac{1}{3}}=6\,volt$

$\frac{1}{r_{eq}}=\frac{1}{3}+\frac{1}{1}+\frac{1}{3}\Rightarrow\frac{3}{5}\Omega$
Steady-state charge on the capacitor
q=CV=($2\mu$)$(6)=12\mu c$
$R_{eq}=\frac{3}{5}+3=\frac{18}{5}\Omega,\,i_{max}=\frac{\varepsilon_{eq}}{R_{eq}}=\frac{6}{\frac{18}{5}}=\frac{5}{3}A$
$R_{eq}C=\frac{18}{5}\times2\Omega=\frac{36}{5}\mu\,sec.$
$\frac{t}{R_c}=\frac{7.2}{\frac{36}{5}}\mu=1$
$i(t)=\frac{\varepsilon_{eq}}{R_{eq}}e^-{\frac{t}{R_{eq}C}}=\frac{5}{3}e^{-1}=\frac{5}{3}\times0.36$
$i=0.6A$
At steady state, voltage across capacitor = 6 V
Therefore, Q = 6 × 2 = 12μC
So, all the options are correct.