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Mathematics LibreTexts

6.3E: The RLC Circuit (Exercises)

  • Page ID
    18281
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    In Exercises [exer:6.3.1} -[exer:6.3.5} find the current in the \(RLC\) circuit, assuming that \(E(t)=0\) for \(t>0\).

    [exer:6.3.1] \(R=3\) ohms;   \(L=.1\) henrys;   \(C=.01\) farads; \(Q_0=0\) coulombs;  \(I_0=2\) amperes.

    [exer:6.3.2] \(R=2\) ohms;   \(L=.05\) henrys;   \(C=.01\) farads’; \(Q_0=2\) coulombs;  \(I_0=-2\) amperes.

    [exer:6.3.3] \(R=2\) ohms;   \(L=.1\) henrys;   \(C=.01\) farads; \(Q_0=2\) coulombs;  \(I_0=0\) amperes.

    [exer:6.3.4] \(R=6\) ohms;   \(L=.1\) henrys;   \(C=.004\) farads’; \(Q_0=3\) coulombs;  \(I_0=-10\) amperes.

    [exer:6.3.5] \(R=4\) ohms;   \(L=.05\) henrys;   \(C=.008\) farads; \(Q_0=-1\) coulombs;  \(I_0=2\) amperes.

    [exer:6.3.6] \({1\over10}Q''+3Q'+100Q=5\cos10t-5\sin10t\)

    [exer:6.3.7] \({1\over20}Q''+2Q'+100Q=10\cos25t-5\sin25t\)

    [exer:6.3.8] \({1\over10}Q''+2Q'+100Q=3\cos50t-6\sin50t\)

    [exer:6.3.9] \({1\over10}Q''+6Q'+250Q=10\cos100t+30\sin100t\)

    [exer:6.3.10] \({1\over20}Q''+4Q'+125Q=15\cos30t-30\sin30t\)

    [exer:6.3.11] Show that if \(E(t)=U\cos\omega t+V\sin\omega t\) where \(U\) and \(V\) are constants then the steady state current in the \(RLC\) circuit shown in Figure [figure:6.3.1} is \[I_p={\omega^2RE(t)+(1/C-L\omega^2)E'(t)\over\Delta},\] where \[\Delta=(1/C-L\omega^2)^2+R^2\omega^2.\]

    [exer:6.3.12] Find the amplitude of the steady state current \(I_p\) in the \(RLC\) circuit shown in Figure [figure:6.3.1} if \(E(t)=U\cos\omega t+V\sin\omega t\), where \(U\) and \(V\) are constants. Then find the value \(\omega_0\) of \(\omega\) maximizes the amplitude, and find the maximum amplitude.

    [exer:6.3.13] \({1\over10}Q''+3Q'+100Q=U\cos\omega t+V\sin\omega t\)

    [exer:6.3.14] \({1\over20}Q''+2Q'+100Q=U\cos\omega t+V\sin\omega t\)

    [exer:6.3.15] \({1\over10}Q''+2Q'+100Q=U\cos\omega t+V\sin\omega t\)

    [exer:6.3.16] \({1\over10}Q''+6Q'+250Q=U\cos\omega t+V\sin\omega t\)

    [exer:6.3.17] \({1\over20}Q''+4Q'+125Q=U\cos\omega t+V\sin\omega t\)