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

The Correct Option is A, B, C, D

\(\frac{x-5}{1}\Rightarrow\,\,x=+6\,,\,i_1=volt,\frac{6-0}{3}=2A\)
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.